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THB *

ELEMENTS OP EUCLI

TI2.

THE FIRST SIX BOOKS,

THE ELEVENTH AND TWELFTH.

THE ERRORS

THE BOOK OF EUCLID'S DATA,

IN LIKE HANKEB COBBBCTSD.

BY ROBERT SIMSON, M. D.

ELEMENTS OP PLANE AND SPHERICAL TRIGONOMB

PHILADELPHIA:
DESILVER, THOMAS & CO.

PHILADELPHIA:

O. SHERMAN & CO., PRINTERS,

19 6t. James street.

(ptA \y^jn^

*-- \^

PREFACE.

Thb opinions of the Moderns concerning the author of the Elements of
Cieometry, which go under Euclid's name, are very dijQferent, and contrary
to one another, reter Ramus ascribes the Propositions, as well as their
Demonstrations, to Theon ; others think the Propositions to be Euclid^s, but
that the Demonstrations are Theon's ; and others maintkin, that all the Pro-
positions and their Demonstrations are Euclid's own. John Buteo and Sir
Henry Savile are the authors of greatest note who assert this last, and
the greater part of geometers have ever sinc# been of this opinion, as they
thought it the most probable. Sir Henry Savile, after the several arguments
. ' he brings to prove it, makes this conclusion (page 13, Prelect^ ''That, ex-

l!^ cepting a very few interpolations, explications, and additions, Theon altered

nothing in Euclid." But, by often considering and comparing together the
Definitions and Demonstrations as they are in the Greek editions we aow.
have, I found that Theon, or whoever was the editor of the present Greek
text, bv adding some things, suppressing others, and mixing his own with
C> Euclid 8 Demonstrations, &d changed more things to the worse than is com-

/ monly supposed, and those not of small moment, especially in the fifth and

ry elevenUi JBooks of the Elements, which this editor has greatly vitiated ; for

I instance, by substituting a shorter, but insufficient Demonstration of the 16th

^ Prop, of the 5th Book, in place of the legitimate one which Euclid had given ;

'") and by takmg out of this Book, besides other things, the good definition which

Eudoxus or Euclid had given of compound ratio, and giving an absurd one
in place of it, in the 5tn Definition of the 6th Book, which neither Euclid)
Archimedes, AppoUonius, nor any geometer before Theon's time, ever made
use of, and of which there is not to be found the least appearance in any of
their writings ; and, as this Definition did much embarrass beginners, and is
quite useless, it is now thrown out of the Elements, and another, which, with-
out doubt, Euclid had given, h put in its proper place among the Definitions
of the 5th Book, by which the doctrine of compound ratios is rendered plain
and easy. Besides, among the Definitions of the 11th Book, there is this,
which is the 10th, viz. ** Equal and similar solid figures are those which are
contained hj siinilar planes of the same number and magnitude," Nw this
Proposition is a Theorem, not a Definition; because the equality of figures
of any kind must be demonstrated, and not assumed ; and therefore, though
this were a true Proposition, it ought to have been demonstrated. But, in-
deed, this Propositioiit which makes the 10th Definition of the 11th Book, is
not true universally, except in the case in which each of the solid angles of
the figures is contained by no more than three plane angles ; fi3r in other
cases, two solid figures may be contained by similar planes of the same num-
ber and magnitude, and yet be unequal to one another, as shall be made evi-
dent in the Notes subjoined to these Elements. In like manner, in the De-
monstration of the 26th Prop, of the 11th Book, it is taken for granted, that
those solid angles are equal to one another, which are contained by plane

«1

IV ' , PREFACE.

aa||les of the same number and magnitude, ^aced in the same order; but
neifner is thi« univef'&ally true, except in the case in which the solid angles
are contained by no more than three plane angles ; nor of this case is there
.idiy Demonstration in the Elements we now have, though it is quite neces-
sary there should be one. Now, upon the 10th Definition of this Book depend
the 25th and 28th Propositions of it; and upon the 25th and 26th depend
other eight, viz., the 27th, 31st, 32d, 33d, 34th, 36th, 37th, and 40th, of the
same B^k ; and the 12th of the 12th Book depends upon the 8th of the
same ; and this 8th, and the Corollary of Proposition 17th and Proposition
18th of the 12lk Book, depend upon the 9th Definition of the 11th Book,
which is not a right definition, because there may be solids contained by the
same number of similar plane figures, which are not similar to one anc^her,
in the true sense of similarity received by all geometers; and all these Pro-
positions have, for these reasons, been insufficiently demonstrated since
Theon*s time hitherto. Besides, there are several other things, which have
nothing of Euclid's accuracy, and which plainly show, that his Elements
have ^en much corrupted by unskilful geometers ; and, though these are
not so gross as the otliers now mentioned, they ought by no means to remain
uncorrected.

Upon these accounts it appeared necessary, and I hope will prove accept-
able, to all lovers of accurate reasoning, and of mathematical learning, to re-
move such blemishes, and restore the principal Books of the Elements to their
original accuracy, as fiir as I was able ; especially since these Elements are
the foundation of a science by which the investigation and discovery of use-
ful truths, at least in mathematical learning, is promoted as far as the limited
powers of the mind allow ; and which likewise is of the greatest use in the
arts both of peace and war, to many of which geometry is absolutely neces-
sary. This I have endeavoured to do, by takmg away the inaccurate and
fiilse reasonings which unskilful editors have put into the place of some of
the genuine Demonstrations of Euclid, who has ever been justly celebrated
as the most accurate of geometers, and by restoring to him those things
which Theon or others have suppressed, and which have, these many ages,
been buried in oblivion.

In this edition, Ptolem/s Proposition concerning a property of quadrilateral
figures in a circle, is added at the end of the sixth Book. Also the Note on
the 29th Proposition, Book 1st, is altered, and made more explicit, and a
more general Demonstration is given, instead of that which was in the Note
on the 10th Definition of Book 11th; besides, the Translation is much amend-
ed by the friendly assistance of a learned gentleman.

To which are also added, the Elements of Plane and Spherical Trigono-
metry, which are commonly taught after the Elements of Euclid.

THE

ELEMENTS OF EUCLID

BOOK I.

DEFINITIONS.

A POINT is that which hath no parts, or which hath no magni-
tude.

A line is length without breadth.

IIL
The extremities of a line are points.

IV,
A straight line is that which lies evenly between its extreme points.

V.
A superficies is that which hath only length and breadth.

VL
The extremities of a superficies are lines.

VU.
A plane superficies is that in which any two points being taken,* the
straight line between them lies wholly in that superficies.

vni.

" A plane angle is the inclination of two lines to one another* in
a plane, which meet together, but are not in the same direc-
tion."

IX,

A plane rectilineal angle is the inclination of two straight lines to
one another, which meet together, but are not in the same straight
line.

A D

* See Notes.

TBe ELEMENTS OF EUCLID.

BOOK I.

N. B. * When sfeveral angles are at ofte point B, any one of them
' is expressed by three letters, of which the letter that is at the
' vertex of the angle, that is, at the point in which the straight lines
' that contain the angle meet one another, is put between the other

* two letters, and one of these two is somewhere upon one of those
' straight lines, and the other upon the other line : Thus the angle
^ which is contained by the straight lines AB, CB, is named the

* angle ABC, or CBA ; that which is contained by AB, DB, is named

* the angle ABD, or DBA; and that which is contained by DB, CB

* is called the angle DBC, or CBD ; but, if th^re be only one angle
< at a point, it may be expressed by the letter placed at that point ;

* as the angle at E.'

X.
When a straight line standing on another
straight line makes the adjacent angles
equal to one another, each of the angles
is called a right angle : and the straight
line which stands on the other is called
a perpendicular to it.

An obtuse angle is that which is greater than a right angle.

XII.
An acute angle is that which is less than a right angle.

xni.

"A term or boundary is the extremity of any thing."

XIV.
A figure is that which is enclosed by one or more boundaries.

XV.
A circle is a plane figure contained by one line, which is called
the circumference, and is such that all straight lines drawn from
a certain point within the figure to the circumference, are equal
to one another:

BOOK I. THB ELEM£NT» OF EUCLID. 7

XVI.

And this point is called the centre of the circle.

xvn. •

A diameter of a circle is a straight line drawn through the centre,
and terminated both ways by the circumference.

XVIII.
A semicircle is the figure contained by a diameter and the part of
the circumference cut off t^ that diameter.

# XIX.

" A segment of a circle is the figure contained by a straight line,
and the circumference it cuts off.'*

XX.

Rectflineal figures are those which are contained by straight lines.

XXI.
Trilateral figures, or triangles^ by three straight lines.

xxn.

Quadrilateral, by four straight lines.

XXffl.

Multilateral figures, or polygons, by more than ibur straight Unes.

XXIV.

Of three sided figures, an equilateral triangle is that which has three
equal sides.

XXV.
An isosceles triangle, is that which has only two sides equal.

XXVI.

A scalene triangle, is that which has three unequal sides.

xxvn.

A right-cmgled triangle, is that which has a right angle.

XXVffl.
An obtuse-angled triangle, is that which has an obtuse angle.

XXIX.
An acute-angled triangle, is that which has three acute angles.

XXX.

Of four-sided figures, a square is that which has all its sides equal,
and all its angles right angles.

>*'

.<r

^^*

THE SLE1IE19T8 OF EUCLID.

BOOK I<

XXXI.

An oblong, is that which has aU its angles right angles, but has not
all its sides equal. •

xxxn.

A rhombus is that which has its sides equat but its angles are
not right angles.

xxxm.

A rhomboid, is that which has its opposite sides equal to ona
another^ but all its sides are not equal, nor its angles right an-
gles.

XXXIV.
All other fi>ur sided figures besides these, are called trapeziums.

XXXV.
Parallel straight lines, are such as are in the same plane, and which
being produced ever so &r both ways^ do not meet.

POSTULATES.

I.
Let it be granted that a straight line may be drawn from any one
pomt to any other point.

II.
That a terminated straight line may be produced to any length in a
straight line.

And that a circle may be described from any centre, at any distance
from that centre.

AXIOMS.

I.
Things whid) are equal to the same are equal to one another.

II.
If equals be added to equals, the wholes are equals.

BOOK I. THfi BLBBIBHTS OF BUCLIO.

in.

If €quals be taken froin eouals, the remainders are equal.

IV.
If equals be added to unequals, the wholes are rniequal.

V.
If equals be taken from unequals, the remainders are uneqaal.

VI.
Things which are double of the same, are equd to one another;

vn.

Thfogs which are hAves of the same, are equal to one another.

vm.

Mi^nitttdes which coincide with one another, that is^ which exaetly
fill the same space, are equal ta one another.

IX.
The whole is greater than its part.
» X.

Two straight lines cannot enclose a space.

XL
AD right angles are equal to one another.

xn.

If a stra^ht Ike meet two straight lines, so as to make the two

** interior angles on the same side of it tiaken together less than
two right angles, these straight lines being continually pro-
duced, shall at length meet upon that side on which are the angles

** which are less than two right angles. See the notes on furop.

-20. of Book I.**

«•
M

«•

**»

TirE e&EMENTS OF CUCLIP.

PROPOSITION I. PROBLEM.

BOOK I.

To describe an equilateral triangle upon a- given finite straight
line.

Let AB be the given straight line ; it is required to describe an
equilateral triangle upon it.

From the centre A, at the distance
AB, describe (S. Postulate.) the cir- /^ /7\\ \

cleBCDv and from the centre B, at ^ // W

the distance BA, describe the circle
ACE; and from the point C» in
which the circles cut one another,
draw the straight lines (2. Post.) CA,
CB to the points A, B ; ABC shall be
an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equaF
(15. Definition.) to AB; and because the point B is the centre of the
circle ACE, BC is equal to BA : but it has been proved that CA hi
equal to AB ; therefore CA, CB are each of them equal to AB ; but
things which are equal to the same are equed to one another; (ist.
Astom.) therefore CA is equal to CB; wherefore CA, AB, BC are
equal to one another ; and the triangle ABC is therefore equilateral,
and it is described upon the given straight line AB. Which was
required to be done.

PROP. IL PROB^

From a given point to draw a straight line equal to a given
straight line.

Let A be the given point, and BC the given straight line ; it is re-
quired to draw fi-om the point A a straight line equal ito BC.

From the point A to B draw (1.
Post.) the straight line AB; and
upon it describe (1. I.) the equilate-
ral triangle DAB, and produce (2.
Post) the straight lines DA, DB,
to E and F; from the centre B, at
the distance BC, describe (9. Post.)
the circle CGH, and from the cen-
tre D, at the distance DG, describe
the circle GKL. AL ^all be equal
toBC.

Because the point B is the centre
of the circle CGH, BC is equal (15.
Def ) to BG ; and because D is the centre of the circle GKL, DL
is equal to DG, and DA, DB, parts c€ them, are equal : therefore
the remainder AL is^ equal to the remainder (3. Ax.) BG ; but It
has been shown, that BC is equal to G, wherefore AL and BC
ore each of them equal to BG ; and things that are equal to the

BOOK. I.

THS £LEMENTS OF EUCLID.

same a]|i|eqaa] to one uiodier; tbereftwe the straight line AL is
equal to^^. Wherefore from the given point A a straight line AL
has been drawn equal to the given straight Hne BC. Which was to
be done.

PROP. m. PROB.

Frok the greater of two given straight lines to cut off a part
equal to the less.

Let AB and C be the two given
straight lines, whereof AB is the
greater. It is required to cut off
firom AB, the greater, a part equal
to C, the less.

From the point A draw (2. 1.) the
straight line AD equal to C; and
from the centre A, and at the dis-
tance AD, describe (3. Post.) the
circle DEF; and because A is the
centre of the chrde DEF, A£ shall be equal to AD ; bnt the straight
line C is likewise equal to AD ; whence AE and C are each of them
equal to AD ; wherefore the straight line AE is equal to (1. Aic.) C,
and from AB, the greater of two straight lines, a part AE has been
cut off equal to C the less. Which was to be done.

PROP. IV- THEOREM.

If two triangles have two sides of the one equal to two sides
of the other, each to each ; and have likewise the angles con-
tained by those sides equd to one another, they shall likewise
have thkr bases, or third sides^ equal; and the two triangles
shall be equal; and their other angles shall be equal, each to
each, viz. those to wjiich the equa) sides are opposite.

Let ABC, DEF be two triangles which l^ave the two sides AB,
AC equal to the two sides DE, DF, each to each, Viz, AB to DE,
and AC to DF ; and the angle A D

BAC equal to the angle EDF,
the base BC shall be equal to
the base EF; and the triangle
ABC to the triangle DEF; and
the other angles, to which the
equal sides are opposite, shall
be equal each to each, viz. the
angle ABC to the angle DEF,
and the angle ACB to DFE. B C E F^

For if the triangle ABC be applied to DEF, so that the point
A may be on D, and the straight line AB upon DE; the point li
shall coincide with the point E, because AB is equal to DE ; and
AB coinciding with DE, AQ shall coincide with DF; because

12 THB BLBMBNTa ^F BUGLID. BOOK I.

the angle BAO is equal to the angle ES^F ; therefore alslfehe point
C shall coincide with the point F, because the straight line AC is
equal to DF : but the point B coincides with the point E ; wherefore
the base BC shall coincide with the base £F, because the point B
coinciding with E, and C with F,' if the base BC does not coincide
with the base EF, two straight lines would inclose a space, which is
impossible. (10. Ax.) Therefore the base BC shall coincide with
the base EF, and be equal to it. Wherefore the whole triangle ABC
shall coincide with the whole triangle DEF, and be equal to it ; and
the other angles of the one shaU coincide with the remainmg angles
of the other and be equal to them, viz, the angle ABC to the angle
DEF, and the angle ACB to DFE, Therefore, if two triangles have
two sides of the one equal to two sides of the other, each to each*
and have likewise the angles contained by thojse sides equal to one
another, their bases shall likewise be equal, and the triangles be equal,
and their other angles to which the equal sides are opposite shall be
equal, each to each. ' Which was to be demonstrated.

PROP. V. THEOB.

The angles at the base of an isosceles triangle are equ^l to
one another : and, if the equal sides be produced, the angles upon
the other side of the base shall be equaL

Let ABC be an isosceles triangle, of which the side AB is equal
to AC, and let the straight lines AB, AC be produced to D and E,
the angle ABC shall be equal to the angle ACB, and the angle CBD
to the angle BCE.

' In BD take any point F, and from AE the greater, cut off AG
equal (3. 1.) to AF, the less, and join FC, GB.

Because AF is equal to AG, and ^B to AC, the two sides FA«
AC, are equal to the two GA* AB, each to each ; and they con*
tain the angle FAG common to the two
triangles AFC, AGB; therefore the base
FC is equal (4. L) to the base (tB, and the
triangle AFC to the triangle AGB; and
the remaming angles of the one are equal
(4. 1.) to the remaining angles of the
other, each to each, ^o which the equal
sides are opposite; viz. the an^e ACF to
the angle ABG, and the angle AFC to the
angle AGB; and because the whole AF is F,
equal to the whole AG, of which the parts
AB, AC, are equal: the remainder BF
shall be equal (8. Ax.) to the remainder jy g

CG; and FC was proved to be equal to
QB; therefore the two sides BF, FC are equal to the two CG,
GB, each to each : and the angle BFC is equal to the angle CGB,
and the base BC is common to the two triangles BFC, CGB ;
wherefore the triangles are equal (4. 1.) and their remaining an-

BOOK I. THB fLBMBMTS OF BOCLID. 13

gles, em^ to each, to which the equal sides, are opposite; there-
fore the angle FBC is equal to the angle GOB, and the angle BCF
to the angle CBG ; and, since it has been demonstrated, that the
whole angle ABG is equal, to the whole ACF, the parts of which,
the angles CBG, BCF are also equal; the remaining angle ABC
is therefore equal to the remaining angle ACB, which are the
angles at the base of the triangle ABC : and it has also been proved
that the angle FBC is equal to the angle GCB, which aire the an-
gles upon the other side of the base. Therefore the angles at the
base, &c. Q. EL D.

Corollary. Hence every equilateral triangle is also equian-
gular.

PROP. VI. THEOR.

If two angles of a triangle be equal to one another, the sidjss
also which subtend, or are opposite to, the equal angles, shall be
equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle
ACB ; and the side AB is also equal to the side AC.

For if AB, be not equal to AC, one of them is greater than the
other; let AB be the greater, and from it cut (3. 1.) off DB equal
to AC the less, and join DC ; therefore, because A

in the triangles DBC, ACB, DB is equal to AC,
and BC common to both, the two sides DB, BC
are equal to the two AC, CB, each to each ; and
the angle DBC is equal to the angle ACB;
therefore the base DC is equal to the base AB,
and the triangle DBC is^ equal to the triangle
(4. 1.) ACB, the less to 'the greater; which is
absurd. Therefore AB is not unequal to AC,
that is, it is equal to it. Wherefore, if two angles /
A;c. Q. E. D, ^ —

Cor. Hence every equiangular triangle is also equilateral.

PROP. Vn. THEOR.

IJpoir the same base, and on the same side of it, there cannot
be two triangles that have their sides which are terininated in
one extremity of the base equal to one another, and likewise
those which are terminated in the other extremity. *

If it be possible, let there be two triangles ACB, ADB, upon the
same base AB, and upon the same side of it, which have tlieir
sides CA, DA, terminated in the extremity A of the base equal
to one another, and likewise their sides, CB, DB, that are terminated
inB.

♦See Note.

THE GLEMENTa OF EOCLIP.

BOOK I.

Join CD; then, in the case in which the CD

vertex of each of the triangles is without the
other triangle, because AC is equal to AD,
the angle ACD is equal (5. 1.) to the angle
ADC : but the angle ACD is greater th^n the
angle BCD ; therefore the angle ADC is great-
er also than BCD ; much more then is the
angle BDC greater than the angle BCD.
Again, because CB is equal to DB, the angle
BDC is equal (5. 1.) to the angle BCD; but^
it has been demonstrated to be greater than
it; which is impossible.

But if one of the vertices, as D, be within the other triangle
ACB; produce AC, AD to E, F; there- E

fore, because AC is equal to AD in the
triangle ACD, the angles ECD, FDC upon
the other side of the base CD are equal
(5. 1.) to one another, but the angle ECD is
greater than the angle BCD ; wherefore the
angle FDC is likewise greater than BCD;
much more then is the angle BDC greater
than the angle BCD. Again because CB
is equal to DB, the angle BDC is equal
(6. 1.) to the angle BDC ; but BCD has been
proved to be greater than the same BCD ; A B

which is impossible. The case in which the
vertex of one triangle is upon a side of the other, needs no demon*
stratlon.

Therefore upon the same base, and on the same side of it, there
cannot be two triangles that have their sides which are terminated
in one extremity of the base equal to one another, and likewise
those which are terminated in the other extremity. CI. E. D.

PROP. Vra. THEOR.

If two triangles have two sides of the one equal to two sides
of the other, each to each, and have likewise their bases equal;
the angle which is contained by the two sides of the one shall
be equal to the angle contained by the two sides equal to them,
of the other.

Let ABC, DEF be
two triangles, having
the two sides AB, AC,
equal to the two sides
DE, DF, each to each,
viz: AB to DE, and
AC to DF; and also
the base BC equal to
the base EF. The an-
gle BAC is equal to
the angle EDF.

D G

C E

BOOI^'I. THE ELIMEMTS OF BUCLID. 15

For, if the triangle ABC be appllM to DEF, sa that the point B
be on E, and the straight line EC upon EF; the point C shall also
coincide with the point F. Because BC is equal to EF ; therefore
BC coinciding with EF, BA and AC shall coincide with ED and
DF ; for, if tlie base BC coincides with the base EF, bat the sides
BA, CA do not coincide with the sides ED, FD but have a different
situation, as EG, FG; tlien upon the same base EF, and upon the
same side of it, theVe can be two triangles that have their sides
which are terminated in one extremity of the base equal to one
another, and lilcewise thdir sides terminated in the other extremity ;
but this is impossible ; (7. 1.) therefore, if the base BC coincides
with the base EF, the sides BA, AC cannot but coincide with the
sides, ED, DF ; wherefore likewise the angle BAC coincides with
the angle EDF, and is equal (8. Ax.) to it Therefore, if two trian-
glesy &c. Q. E. D,

PROP. IX. PROB.

To bisect a given rectilineal angle, that is, to divide it into two
equal angles.

Let BAC be the given rectilineal angle, it is required to bisect it.

Take any point D in AB, and from AC cut (3. 1.) off AE equal
to AD ; join DE, and upon it describe (1« 1.) A

an equilateral triangle DEF ; then join AF ;
the straight line AF bisects the angle BAC.

Because AD is equal to AE, and AF is
common to the two triangles DAF, EAF;
the two sides DA, AF, are equal to the two
sides, EA, AF each to each ; and the base
DF is equal to the base EF ; therefore the
angle DAF is equal (8. 1.) to the angle EAF ;
wherefore the given rectilineal angle BAC is
bisected by the straight linCj AF, whidi was ^
to be done. ^

PROP. X. PROB.

To bisect a given finite straight line, that is^ to divide it into
two equal parts.

Let AB be the given straight line : it is required to divide it inta
two equal parts.

Describe (1. 1.) upon it an equilateral triangle ABC, and bisect!
(9. 1.) the angle ACB by the straight line CD. AB is cut into two
equal parts in the point D.

THB BLBMEHTS OF EUGUD.

BOOK, t*

Because AC is equal to CB^ and CD com-
mon to the two triangles ACD, BCD; the
two sides AC, CD are equal to BC, CD, each
to each ; and the angle ACD is equal to the
angle BCD; therefore the base AD is equal
to the base (4. !•) DB and the straight line
AB is divided into two equal parts in the point
D. Which was to be done.

PROP. XI. PROB.

To draw a straight line at right angles to a given straight
line, from a given point in the same*

Let AB be a ^ven straight line, and C a point given in it : it is re-
quired to draw a straight line from the point C at right angles to AB.*^

Take any point D in AC, and (3. 1.) make CE equal to CD, and
upon D£ describe (1. 1.) the equilateral F

triangle, DFE, and join PC; the
straight line FC drawn from the given
point C is at right angles to the given
straight line AB. .

Because DC is equal to CE, and FC
common to the two triangles DCF,
ECF ; the two sides, DC, OF, are equal
to the two EC, CF, each to each ; and
the base DF is equal to the base EF ; therefore the angle DCF is
equal (8. 1.) to the angle ECF ; and they are adjacent angles; But,
when the adjacent angles which one straight line makes with another
stra^ht line are equal to one another, each of them is called a right
(10. Def. 1.) angle ; therefore each of the angles DCF, ECF, is a
right angle. Wherefore, from the given point C, in the given
straight line AB, FC has been drawn at right angles to AB. Which
was to be done. .

Cor. By help of this problem, it may be demonstrated, that two
straight lines cannot have a coibmon segment

If it be possible, let the two straight lines ABC, ABD have the s^-
ment AB common to both of them. From the point B draw BE at
right angles to AB; and because ABC is a straight line, the angle

CBE is equal (10. Def. 1.) to the
angle EBA ; in the same manner,
because ABD is a straight line, the
angle DBE is equal to the angle
EBA; wherefore the angle DBE
is equal to the angle CBE, the less
to the greater ; which is impossible;
therefore two straight lines cannot
have a common segment.

• See Note.

BOO£ I.

TBS ELfiMBHTS OF SUCLfO.

PROP. Xn. PROB.

To draw a straight line perpendicular to a given straight line
of an unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced to
any length both ways, and let C be a point without it It is re-
quired to draw a straight line per- C
pendicular to AB from the point C.

Take any point D upon the other
side of AB, and from the centre C,
at the distance, CD, describe (3. Post)
the circle FDG meeting AB in P, G;

and, bisect (10. 1.) FG in H, and join . ^

CF, CH, CG; the straight line CH,^ *
drawn from the given point C, is per-
pendicular to the given straight line AB.

Because FH is equal to HG, and HC, common to the two trian-
gles FHC, GHC, the two sides FH, HC are equal to the two GH,
HC» each to each ; and the base CF is equal (15 Def. 1.) to the base
CG ; therefore the angle CHF is equal (8. 1.) to the angle CHG; and
they are adjacent angles; but when a straight line standing on a
straight line makes^ the adjacent angles equal to one another, each
of them is a right angle, and the straight line which stands upon the
other is called a perpendicular to it ; therefore from the given point
C a perpendicular CH has been drawn to the given straight line AB.
Which was to be done.

PROP. XnL THEOR.

The angles which one. straight line makes with another upon
the one side of it, are either two right angles, or are together
equal to two right angles.

Let the straight line AB make with CD, upon one side of it, the
angles CBA, ABD ; these are either two right angles, or are together
equal to two right angles.

For, if the angle CBA be equal to ABD, each of them is a right

A E A

D B C

(def 10.) angle; but, if not, from the point B draw BE at right an-
gles (11. 1.) to CD; therefore the angles CBE, EBD, are two right
angles ; (def. 10.) and because CBE is equal to the two angles CBA,

18 THE fibBMBNTS OF BUCLIO. BOOK I.

ABE, together, add the angle EBD to each of these equals; there-
fore the angles CBE, EBD are (2 Ax.) equal to the three angles CBA,
ABE, EBD« Again, because the angle DBA is equal to the two an-
gles DBE, EBA, add to these equals the angle ABC ; therefore the
angles DBA, ABC are equal to the three angles DBE, EBA, ABC ;
but the angles CBE, EBD have been demonstrated to be equal to
the same three angles ; and things that are equal to the same are
equal (1. Ax.) to one another; therefore the angles CBE, EBD are
equal to the angles DBA, ABC: but CBE EBD are two right angles:
therefore DBA, ABC are together equal to two right angles. Where-
fore, when a' straight line, &c. d. K. D.

PROP. XIV. THEOR.

If, at a point in a straight line, two other straight lines, upon
the opposite sides of it, make the adjacent angles together equal
to two right angles, these two straight lines shall be in one and
the same straight line.

At the pcnnt B in the straight line
AB, let the two straight lines BC, BD
upon the opposite sides of AB make
the adjacent angles ABC, ABD equal
together to two right angles, BD is
in the same straight line with CB.

For, if BD be not in the same
straight line with CB, let BE be in
the same straight line with it; there-
fore, because the straight line AB ^
makes angles witir~the straight line CBE, upon one side of it, the
angles ABC, ABE are together equal (13. 1.) to two right angles ;
but the angles ABC, ABD are likewise together equal to two right
angles ; therefore the angles CBA, ABE are equal to the angles CBA,
ABD : take away the common angle ABC, the remaining angle ABE
is equal (3. Ax.) to the remaining angle ABD, the less to the greater,
which is impossible ; therefore BE is hot in the same straight line
with BC. And, in like manner, it may be demonstrated that no
other can be in the same straight line with it but BD, which there-
fore is in the same straight line with CB. Wherefore, if at a point,
&c. Q. E. D.

PROP. XV. THEOR.

If two straight lines cut one another, the vertical or opposite
angles shall be equal

Let the two straight lines AB, CD cut one another in the point E;
the angle AEC ^hall be equal to the angle DEB, and CEB to
AED.

BOOK U

TH0 JBLBMfiirro' OF EUGUD.

Because the straight line AE makes
with CD the angles CEA, AED, these C
angles are together equal (13. 1.) to
two right angles. Again, because the

straight line DE makes with AB the A
angles AED, DEB, these also are to-
gether equal (13. 1.) to two right an-
gles; and CEA, AED have been de- D
monstrated to be equal to two right angles ; wherefore the angles
CEA, AED are equal to the angles AED, DEB. Take away the
common angle AED, and the remaining angle CEA is equal (3. Ax.)
to the remaining angle DEB. In the same manrier it can be de-
monstrated that the angles CEB, AED are equal. Therefore, if two
straight lines, &.c. Q. E. D.

Cor. 1. From this it is manifest, that if two straight lines cut one
another, the angles they make at the point where they cut, are to-
gether equal to four right angles.

Cor. 2. And consequently that all the angles made by any num-
ber of lines meeting in one point, are together equal to four right
angles.

PROP. XVI. THEOR.

If one side of a triangle be produced, the exterior angle is
greater than either of the interior opposite angles.

Let ABC be a triangle, and let its side BC be produced to D, the
exterior angle ACD is greater than either of the interior opposite
angles, CBA, BAC. ^

Bisect (10. 1.) AC in E, j»in BE
and produce it to F, and make EF
equal to BE ; join also FC, and pro-
duce AC to G.

Because AE is equal to EC, and
BE to EF; AE, EB are equal to
CK, EF^ each to each ; and the an-
gle AEB is equal (15. 1.) to the an»
gle CEF, because they are opposite
verticsd angles ; therefi;>re the base
AB is equal (4. 1.) to the base CF,
and the triangle AEB to the triangle
CBIF, and the remaining angles, to
the remaining angles, each to each,
to which the equal sides are opposite ; wherefore the angle BAE is
equal to the angle ECF; but the angle ECD is greater than the an-
gle ECF ; therefore the angle ACD is greater than BAE : in the same
manner, if the side BC be bisected, it may be demonstrated that the
angle BCO, that is, (15. 1.) the angle ACD, is greater than the angle
ABC. Therefore, if one side, &c. Ct E. D.

20 THE GLEMBNTS OP EUCLID. BOOK I.

PROP. XVn. THEOR.

Any two angles of a triangle are together less than two right
angles.

Let ABC be any triangle ; any two of
its angles together are less than two right
angl^.

Produce BC to D ; and because ACD is
the exterior angle of the triangle ABC, ACD
is greater (16. 1.) than the interior and op*
posite angle ABC ; to each of these add .
the angJe ACB; therefore the angles ACD, /
ACB are greater than the angles ABC, B ' C D

ACB; but ACD, ACB are together equal (13. 1.) to two right angles;
therefore the angles ABC, BCA are less than two right angles. In
like manner, it may be demonstrated, that BAC, ACB, as jJso CAB,
ABC, are less than two right angles. Therefore any two angles, &c»
Q. £. D.

PROP. XVIII. THEOR.

The greater side of every triangle is opposite to the greater
angle.

Let ABC be a triangle, of which the A

side AC is greater than the side AB ;
the angle ABC, is also greater than'
the angle BC^.

Because AC is greater than ABy /
make (3. 1.) AD equal to AB, and /
join BD ; and because ADB is the ex- '^
tenor angle of the triangle BDC, it is B ^

greater (16. 1.) than the interior and opposite angle DCB; but ADB
is equal (5. 1.) to ABD, because the side AB is equal to the side ADf
therefore the angle ABD is likewise greater than the angle ACB;
wherefore much more is the angle ABC greater than ACB. There-
fore the greater side, &c. Gl. E. D.

,««

PROP. XIX. THEOR.

The greater angle of every triangle is subtended by the greater
side, or has the greater side opposite to it

Let ABC be a triangle, of which the angle ABC is greater than the
angle BCA; the side AC is likewise greater than the side AB.

BOOK I.

THE BLBMEHTS OP EUCUD.

For, if it be riot greater, AC must
either be. equal to AB, or less than it;
it is not equal, because then the angle
ABC would be equal (5. 1.) to the an-
gle ACB ; but it is not ; therefore AC is
not equal to AB ; neither is it less ; be-
cause then the angle ABC would be _
less (18, 1.) than the angle ACB ; but B C

it is not; therefore the side AC is not less than AB ; and it has been^
shown that it is not equal to AB; therefore AC is greater than AB.
Wherefore the greater angle, &c. €1. E. D.

PROP. XX. THBOR.

Any two sides of a triangle are together greater than the
third side. * .

Let ABC be a triangle ; any two sides of it together are greater
than the third side, viz. the sides BA, AC greater than the side BC ;
and AB, BC greater than AC ; and BC, CA greater than AB.

Produce BA to the pomt D, and D

make (3. 1.) AD equal to AC ; and join
J3C. A

Because DA is equal to AC, the an-
gle ADC is likewise equal (5. 1.) to
ACD; but the angle BCD is greater
than the angle ACD ; therefore the an-
gle BCD is greater than the angle ADC ; ^
and because the angle BCD of the triangle DCB is greater than
its angle BDC, and that the greater (19. 1.) side is opposite to the
greater angle : therefore the side DB is greater than the side BC ;
but DB is equal to BA and AC ; therefore the sides BA, AC are
greater than BC. In the same manner it may be demonstrated, that
the sides AB, BC are greater than CA, and BC, CA greater than
AB. Therefore any two sides, &c. d. E. D.

PROP. XXI. THEOR.

«
Ir, from the ends of the side of a triangle, there be drawn
two straight lines to a point within the triangle, these shall be
less than the other two sides of the triangle, but shall contain a
greater angle. *

Let the two straight lines BD, CD be drawn from B, C, the ends
of the side BC of the triangle ABC, to the point D within it ; BD
and DC are less than the other two sides BA, AC of the triangle^
but contain an angle BDC greater than the angle BAC.

Produce BD to E; and because two sides of a triangle are
greater than the third side, the two sides BA, AE of the triangle

* See Notes.

THB BLEMEZITS OF BOCLID.

B09X I*

ABE are greater than BE. To each of these add EC ; therefore the

sides BA, AC are greater than BE, A

EC : agaitt, because the two sides 0£2,

ED of the triangle CED are greater

than CD, add DB to each of these;

therefore the sides CE, EB are greater

than CD, DB ; but it has been shown

that BA, AC are greater than BE, EC;

much more then ate BA, AC greater

than BD, DC.

Again, because the exterior angle B
of a triangle is greater than the interior and opposite angle, the
exterior angle BDC of the triangle CDE is greater than CED 5 for
the same reason, the exterior angle CEB of the triangle ABE is
greater than BAC ; and it has been demonstrated that the angle
BDC is greater than the angle CEB; much more then is the angle
BDC greater than the angle BAC. Therefore, if from the ends of,
&c Q. E. D.

PROP. XXIL PROS.

To make a triangle of which the sides shall be equal to three
given straight lines, but any two whatever of these must be
greater thaathe third (20. 1.)*

Let A, B, C be the three given straight lines, of which any two
whatever are greater than the third, viz. A and B, greater than C ;
A and C greater than B; and B and-C than A. It is required to
make a triangle of which the sides shall be equal to A, B, C, each to
each.

Take a straight line DE terminated at the point D, but unli-
mited towards E, and make

(3. 1.) DF equal to A, FG to ^ >. j^

B, and GH equal to C ; and
from the centre F, at the dis-
tance FD, describe (3 Post.)
the circle DKL: and from the
centre G, at the distance GH,
describe (3. Post.) another
circle HLK: and join KF,
KG; the triangle EFG has
its sides equal to the three
straight lines A, B, C.

Because the point F is in the centre of the circle BKL, FD is
equal (16. Def ) to FK ; but FD is equal to the straight line A ;
therefore FK is equal to A : again, because G is the centre of the
circle LKH, GH is equal (15. Def.) to, GK ; but GH is equal to
C ; therefore also GK is equal to C ; and FG is equal to B ; there-
fore the three straight lines KF, FG, GH, are equal to the three
A, B, C ; and therefore the triangle ICFG has its three sides KF,

* Sec Note.

BOOK U THE BLEIIEUT* OF EUCLID. S8

FG, OK equal to three given straigbt lines, A, B, C. Which was to
be done.

PROP. XXm. PROB.

At a given point in a given straight line, to make a rectilineal
angle equal to a given rectilineal angle.

Let AB be the given straight Ihae, and A the given point in it,
and DCE the given rectilineal angle; It is required to make an
angle at the given point A in C A

the given straight line AB, that
shall be equal to the given recti-
lineal angle DCE.

Take in CD, CE, any points
D, E, and join DE, and make
(22. 1.) the triangle AFG, the
sides of which shall be equal
to the three straight lines D'
CD, DE, CE, so that CD be
equal to AP ; CE to AG : and
DE to FG; and because DC,
CE are equal to FA, AG, each to each, and the base DE to the base
FG; the angle DCE is equal (8. 1.) to the angle FAG. Therefore^
at the given point A in the given straight line AB, the angle FAG
is made equal to the given rectilineal angle [DCE. Which was to
be done.

PROP. XXIV. THEOR.

If two triangles have two sides of the one equal to two sides
of the other, each to each, but the angle contained by the two
sides of one of thetn greater than the angle contained by the
two sides equal to them, of the other ; the base of that which
has the greater angle shall be greater than the base of the
other.*

Let ABC, DEF be two triangles which have the two sides AB,
AC equal to the two D£^, DF, each to each, viz. AB equal to DE,
and AC to DF ; but the angle BAC greater than the ang^le EDF ;
the base 6C is also greater than the base EF.

Of the two sides DE, DF, let DE be the side which is not greater
than the other, and at the point D, in the straight line DE, make
(23. L) the angle EDG equal to the angle BAC ; and make DG equal
(8. 1.) to AC or DF, and join EG, GF.

Because AB is equal to DE, and AC to DG, the two sides BA,
AC are equal to the two ED, DG, each to each, and the angle

* See Note*

THE ELEMENTS OF EUCLID.

BOOK 1.

BAG is equal to the angle
EDG; therefore the base
BC is equal (4. 1.) to the
base EG; and because
DG is equal to DF, the
angle DFG is equal (5. 1.)
to the angle DGF; but
the angle DGF is greater
than the angle EGF;
therefore the angle DFG
is greater than EGF ; and

much more is the angle EFG greater than the angle EGF ; and be-
cause the angle EFG of the triangle EFG is greater than its angle
EGF, and that the greater (19. 1.) side is opposite to the greater
angle ; the side EG is therefore greater than the side EF ; but EG
is equal to EG ; and therefore also EG is greater than EF« There-
fore, if two triangles, dec. Q.. E. D.

PROP. XXV. THEOR.

If two triangles have two sides of the one equal to two
sides of the other, each to each, but the base of the one greater
than the base of the other ; the angle also contained by the sides
of that which has the greater base, shall be greater than the
angle contained by the sides equal to them, of the other.

Let ABC, DEF be two triangles which have the two sides AB,
AG equal to the two sides DE, DF, each to each, viz. AB equal to
DE, and AG to DF ; but the base GB is greater than the base EF ;
the angle BAG is likewise greater than the angle EDF.

For, if it be not greater, it must either be equal to it, or less ;
but the angle BAG is not equal to the angle EDF, because then
the base BG would be equal A D

(4. 1.) to EF; but it is not;
therefore the angle BAG is not
equal to the angle EDF, nei-
ther is it less ; because then
the base BG would be less (24.
1.) than the base EF ; but it is
not ; therefore the angle BAG
is not less than the angle EDF

and it wats shown that it is not equal to it ; therefore the angle BAG
Is greater than the angle EDF. Wherefore, if two triangles, &c.

a. E. D.

PROP. XXVI. THEOR.

If two triangles have two angles of one equal to two angles
of the other, 'each to each ; and one side equal to one side, viz.
either the sides adjacent to the equal angles, or the sides opposite

BOOK 1.

THE ELEMENTS OV EUCLID.

to equal angles in each ; then shall the other sides be equal,
each to each; and also the third angle of the one to the third
angle of the other.

Let ABC, DEF be two triangles which have the angles ABC, BCA
equal to the angles DEF, EPD, viz. ABC to DEF, and BCA to EFD;
also one side equal to one side ; and first let those sides be equal
which are adjacent to the A D

angles that are equal in
the two triangles, viz*
BC to EF; the other G
sides shall be equal, e^ch
to each, viz. AB to DE,
and AC to DF : and the
third angle BAC to the
third angle EDF.

For, if AB be not equal B C E F

to DE, one of them must be the greater. Let AB be the greater of
the two, and make B6 equal to DE and join GC ; therefore, because
BG is equal to DE, and BC to EF, the two sides GB, BC are eual
to the two DE, EF, each to each; and the angle GBC is equal to the
angle DEF; therefore the base GC is equal (4. I.) to the base DF,
and the triangle GBC to the triangle DEF, and the other angles to
the other angles, each to each, to which the equal sides are opposite;
therefore the angle GCB is equal to the angle DFE; but DFE is, by
the hypothesis, equal to the angle BCA; wherefore also the angle
BCG la equal to the angle BCA, the less to the greater, which is
impossiUe ; therefore AB is not unequal to DE, that is, it is equid
to it, and BC is equal to EF ; therefore the two AB, BC are equal to
the two DE, EF, each to each ; and the angle ABC is equal to the
angle DEF; the base therefore AC is equal (4. 1.) to the base DF,
and the third angle BAC to the third angle EDF.

Next let the sides ^

which are opposite to
equal angles in each tri-
angle be equal to one
another, viz. AB to DE ;
likewise in this case, the
other sideis shall be
equal, AC to DF, and
BCtoEF; and also the
third angle BAC to the
third EDF.

H C

For, if BC be not equal to EF, let BC be the greater of them,
and make BH equal to EF, and join AH ; and because BH is equal
to EF, and AB to DE, the two AB BH are equal to the two DE,
EF each to each; and they contain equal angles; tiierefore the
base AH is equal to the base DF, and the triangle ABH to the
triangle DEF, and the other angles shall be equal, ea^ch to each,
to which the equal ndes are opposite; therefore the angle BHA

THE BLEMBNTS OF EUCLID.

BOOK 1.

is equal to the angle EFD ; but EFD is equal to the angle BCA ;
therefore also the angle BHA is equal to the angle BCA, that is, the
exterior angle BHA of the triangle AHC is equal to its interior
and opposite angle BCA; which is impossible; (16. 1.) wherefore
BC is not unequal to £F, that is, it is equal to it ; and AB is equal
to DE; therefore the two AB, BC are equal to the two DE, EF, each
to each; and t\^y contain equal angles; wherefore the base AC is
equal to the base DF, and the third angle BAC to the third angle
EDF. Therefore, if two triangles, &c. O. E. D.

PROP. XXVn. THEOR.

If a straight line falling upon two other straight lines makes
the alternate angles equal to one another, these two straight lines
shall be parallel.

Let the straight line EF, which ^Is upon the two straight lines
AB, CD make the etlternate angles AEF, EFD equal to one another;
AB is parallel to CD.

For, if it be not parallel, AB and CD being produced shall meet
either towards B, D, or towards A, C ; let them be produced and
meet towards B, D, in the point G ; therefore GEFis a triangle, and
its exterior angle AEF is greater (16. 1.) than the interior and op*
posite angle J^G; but it is
also equal to it, which is im-
possible; therefore AB and A E/ h
CD being produced do not

meet towards B, D. In like / ^^ G

manner it may be demonstrat- C
ed that they do not meet to-
wards A, C ; but those striaight
lines which meet neither way,

though produced ever so far, are parallel (35. def.) to one another.
AB therefore is parallel to CD. Wherefore, if a straight line, &c.
Q. E. D.

PROP. XXVra. THEOR.

If a straight line falling upon two other straight lines makes
the exterior angle equal to the interior and opposite upon the
same side of the line; or makes the interior angles upon the same
side together equal to two right angles ; the two straight lines
shall be parallel to one another.

Let the straight line EF, which Ms E
upon the two straight lines AB, CD,
make the exterior angle EGB equal to x g

the interior and opposite angle GHD A N: ■ B

upon the same side ; or make the inte-
rior angles on the same side BGH,

GHD together equal to two right an- ^ \ D

gles ; AB is parallel to CD. ii

Because the angle EGB is equal to

Boot t. fHS ELEMENTS OF EUCtlD. 2t

the angle GHD, and the angle EGB, equal (15. 1.) to the angle
AGH, the angle AGH is equal to the angle GHD ; and they are the
alternate angles; therefore AB is parallel (27. 1.) to CD. Again,
because the angles, BGH, GHD are equal (by hyp.) to two right
angles; and that AGH, BGH are also equal (13. 1.) to two right
angles ; the angles AGH, BGH are equal to the angles BGH, GHD:
take away the common angle BGH ; therefore the remaining angle
AGH is equal to the remaining angle GHD ; and they are alternate
angles : therefore AB is parallel to CD. Wherefore, if a straight
line, &c. Q. E. D.

PROP. XXIX. THEOR.

If a straight line fall upon two parallel straight lines, it makes
the alternate angles equal to one another; and the exterior an^le
equal to the interior and opposite upon the same side ; and like-
wise the two interior angles upon the same side together equal to
two right angles.*

Let the straight line EF fall upon the parallel straight lines AB,
CD ; the alternate angles, AGH, GHD are equal to one another; and
the exterior angle EGB is equal to the interior and opposite upon the
same side GHD, and the two interior
angles BGH, GHD upon the same
side are together equal to two right
angles . ^ V^; g

For if AGH be not equal to GHD,
one of them must bef greater than
the other; let AGH be the greater; _
and because the angle AGH is great- ^
er than the angle GHD, add to each
of them the angle BGH, therefore
the angles AGH, BGH are greater ^

than the angles BGH, GHD ; but the angles AGH, BGH are equ^l
(13. 1.) to two right angles; therefore the angles BGH, GHD are
less than two right angles; but those straight lines which, with
another straight line falling upon them, make the interior angles on
the same side less than two right angles, do meet (12. ax.) * together
if continually produced ; therefore the straight lines AB, CD, if pro-
duced far enough, shall meet ; but they never meet, since they are
parallel by the hypothesis ; therefore the angle AGH is not unequal
to the angle GHD, that is, it is equal to it ; but the angle AGH is
equal (15. 1.) to the angle EGB; therefore likewise EGB is equal to
GHD ; add to each of these the angle BGH ; therefore the angles
EGB, BGH are equal to the angles BGH, GHD ; but EGB, BGH are
equal (13. 1.) to two right angles; therefore also BGH, GHD are
equal to two right angles. . Wherefore, if a straight lin^ &c. Q,.
E.D.

* Siee the Notes to this Proposition.

88 THB fi&EMBirra OF bhcud. iiook i.

PROP. XXX. THEOR.

Straight lines which are parallel to the same straight line are
parallel to one another.

Let AB, CD, be each of them parallel, to £F, ABis also parallel
to CD.

Let the straight line GHK cut AB» EF, CD ; and because GHI^
cuts the parallel straight lines AB, EF, the
angle AGH is equal (29. 1.) to the angle
GHF. Again, because the straight line GK a
cuts the parallel straight lines EF, CD, the
angle GHF is equal to (29. I.) the angle
GKD ; and it was shown that the angle AGK E
is equal to the angle GHF ; therefore also
AGK is equal to GKD; and they are alter- C
nate angles ; therefore AB is parallel (27. 1 .)
to CD. Wherefore straight lines, &c.
Q.E.D.

PROP. XXXI. PROB.

To draw a straight line through a given point parallel to a
given straight line.

Let A be the given point, and BC the given straight line ; it is
required to draw a straight line through E A F

the point A, paraUel to the straight line ""
BC.

In BC take any point D, and join AD ;

and at the point A, in the straight line AD, ^ ^ ^

make (23. 1.) the angle DAE equal to the

angle ADC ; and produce ^he straight line EA to F.

Because the straight line AD, which meets the two straight lines
BC, EF, makes the alternate angles EAD, ADC equal to one ano-
ther, EF is parallel (27. 1.) to BC. Therefore the straight line EAF is
drawn through the given point A parallel to the given straight line
BC. Which was to be done.

PROP. XXXII. THEOR.

If a side of any triangle be produced, the exterior angle is
equal to the two interior and c^posite angles; and the three in-
terior angles of every triangle are equal to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced
to D; the exterior angle ACD is equal to the two interior and op-
posite angles CAB, ABC ; and the three interior angles of the

BOOK I.

THB aUBMBMTS OP RUCLIO.

ttfi

trian^, viz. ABC, BCA, CAB, are together equal to two right
angles.

Through the point C draw A

C£ parallel (31. 1.) to the ^A .E

straight line AJB; and because
AB is parallel to C£ and AC
itieets them, the ahemate an-
gles BAC, ACE are equal (20.
1.) Again; because A is pa- ^ CD

rallel to CB, and BD Ms upon them, the exterior angle ECD is
equal to the interior and opposite angle ABC ; but the angle ACE
was shown to be equal to the angle BAC ; therefore the whole ex-
terior angle ACD is equal to the two interior and opposite angles
CAB, ABC ; to these equals add the angle ACB, and the angles
ACD, ACB are equal to the three angles CBA, BAC, ACB ; but the
angles ACD, ACB are equal (13. 1.) to two right angles: therefore
also the angles, CBA, BAC, ACB are equal to two right angles.
Wherefore, if a side of a triangle, &c. Q. E. D.

Cor. 1. All the interior angles of any
rectilineal figure, together with four right
angles, are equal to twice as many right
angles as the figure has sides.

For any rectilineal figure ABCDE can
be divided into as many triangles as the
figure has sides, by drawing straight lines
firom a point F within*the ^gure, to each of
its angles. And, by the preceding proposi-
tion, all the angles of these triangles are A B
equal to twice as many right angles as there are triangles, that is,
as there are sides of the figure ; and the same angles are equal to
the angles of the figure, together with the angles at the point F,
which is the common vertex of the triangles : that is, (2 Cor. 16. 1.)
together with four right angles. Therefore all the angles of the
figure, together with four right angles, are equal to twice as many
right angles as the figure has sides.

CoR. 2. All the exterior angles of any rectOineal figure, are to-
gether equal to four right angles.

Because every interior angle ABC,
with its adj&cent exterior ABD, is
equal (13. 1.) to two right angles;
therefore all the interior, together
with all the exterior angles of the
figure, are equal to twice as many
right angles as there are sides of the
figure ; that is, by the foregoing co- jy
rollary, they are equal to all the in- ^ i^

tenor angles of the figure, together
with four right angles ; therefore all the exterior angles are equal to
four right angles*

T0K ELEMENTS OF EUCLID* BOOK 1.

PROP. XXXra. THEOR.

The straight lines which join the extremities of two equal and
parallel straight lines, towards the same parts, are also them-
selves equal and parallel.

Let AB, CD be equal and parallel straight ^ q

lines, and joined towards the same parts by
the straight lines AC, BD; AC, BD are also
equal and parallel.
> Join BC ; and because AB is parallel to

CD ; and BC meets them, the alternate an- _^__

gles ABC, BCD are equal (29. 1.) ; and be- ^ D

cause AB is equal to CD, and BC common to the two triangles ABC,
DCB, the two sides AB, BC are equal to the two DC, CB ; and the
angle ABC is equal to the angle BCD ; therefore the base AC is equal
(4. L) to the base BD, and the triangle ABC to the triangle BCD,
and the other angles to the other angles, (4. 1.) each to each, to which
the equal sides are opposite : therefore the angle ACB is equal to the
angle CBD ; and because the straight line BC meets the two straight
lines AC, BD, and makes the alternate angles ACB, CBD equal to
one another, AC is parallel (27. 1.) to DB; and it was shown to be
equal to it. Therefore straight lines, &c. Q. E. D.

PROP. XXXIV. THEOR.

The opposite sides and angles of parallelograms are equal to
one another, and the diameter bisects them, that is, divides them
into two equal parts.

N. B. S. parallelogram is a four aided figure, of which the oppo-
site sides are parallel; and the diameter is the straight line joining
two of its opposite angles.

Let ACDB be a parallelogram, of which BC is a diameter ; the
opposite sides and angles of the figure are equal to one another ; and
the diameter BC bisects it.

Because AB is parallel to CD, and A B

BC meets them, the alternate angles
ABC, BCD are equal (29. 1.) to one ano-
ther ; and because AC is parallel to BD,
and BC meets them, the alternate angles
ACB, CBD are equal (29. 1.) to one ano-
ther; wherefore the two triangles ABC,
CBD have two angles ABC, BCA in one, equal to two angles
BCD, CBD in the other, each to each, and one side BC common
to the two triangles, which is adjacent to their equal angles;
therefore their other sides shall, be equal, each to each, and the
third angle of the one to the third angle of the other, (26. 1.)
viz. the side AB to the side CD, and AC to BD, and the angle
BAC equal to the angle BDC; and because the angle ABC is

JOOK I.

THB BLEMfiNTfl OF BUCLID.

equal to the angle BCD, and the angle CBD to the angle ACB, the
whole angle ABD is equal to the whole angle ACD : and the angle
BAC has been shown to he equal to the angle BDC ; therefore the
opposite sides and angles of parallelograms are equal to one another ;
also, their diameter bisects them ; for AB being equal to CD, and
BC common, the two AB, BC are equal to the two DC, CB» each to
each ; and the an^e ABC is equal to th^ angle BCD ; therefore the
triangle ABC is equal (4. 1 .) to the triangle BCD, and the diameter
BC divides the parallelogram ACDB into two equal parts. Q. E. D.

PROP. XXXV. THEOR.

Parallelograms upon the same base, and between the same
parallels, are equal to one another.'*^

Let the parallelograms ABCD, EBCF be upon the same base BC,'
and between the same parallels AF, BC; the parallelogram ABCD
shall be equal to the parallelogram EBCF.t

If the sides AD, DF of the parallelo- A D

grams ABCD, DBCF opposite to the base
BC be terminated in the same point D ; it
is plain that each of the parallelograms is
double (34. 1.) of the triangle BDC; and
they are therefore equal to one another.

But, if the sides AD, EF, opposite to B i C
the base BC of the parallelograms ABCD, EBCF, J^e not terminated
in the same point ; then, because ABCD is a pa' /allelogram, AD is
equal (34. 1.) to BC ; for the same reason EF is 4^quai to BC ; where-
fore AD is equal (1. Ax.) to EF ; and DE is crj/j^jnon ; therefore the
whole, or the remainder AE, is equal (2. or h^ ^x.) to the whole,
or the remainder DF ; AB also is equal totfjyQ . ^nd the two EA
AB are therefore equal to the two FDjVj^q each to each; and

D E

B C

the exterior angle FDC is equal (19J
fore the base EB is equal to the b;
equal (4. 1.) to the triangle FDC ; i
trapezium, ABCF, and from the s
EAB ; the remainders therefore
parallelogram ABCD is equal to t
fore, parallelograms upon the samv

» See Note.

B C

1.) to the interior EAB : there-
ise FC, and the triangle EAB
iake the triangle FDC from the
fame trapezium take the triangle
kre equal, (3. Ax.) that is, the
[he parallelogram EBCF. There-
base, &.C. €1. E. D.

t See the 2d *nd 3d figure*.

J..

THE BLBMENTS OF EUCLID.

BOOK !•

PROP. XXXVI. THEOR.

Parallelograms upon equal bases, and between the same
parallels, are equal to one another.

Let ABCD, EFGH be pa- A
rallelograms upon equal bases
BC, PG, and between the same \
parallels AH, BG; the paral- |
lelogram ABCD is equal to
EFGH.

Join BE, CH ; and because ______

BC is equal to FG, and FG B ~ C F G

to (34. 1.) EH, BC is equal to EH ; and they are parallels, and joined
towards the same parts by the straight lines BE, CH ; but straight
]ines which join equal and parallel straight lines towards the same
parts, are themselves equal and parallel ; (33. 1.) therefore EB, CH
axe both equal and parallel, and EBCH is a parallelogram ; and it is
equal (35. 1.) to ABCD, because it is upon the same base, BC, and
betweel^e same parallels BC, AD : for the like reascm, the paral-
lelogram ilifGH is equal to the same EBCH ; therefore also the pa-
rallelogram 'ABCD is equal to EFGH. Wherefore parallelograms,
&c. Q. E. D. X

\ PROP. XXXVn. THEOR.

Tbiangles up^ the same base, and between the same paral-
lels, are equal to Vie another.

Let the trianglesliBC, DBC be upon the same base BC, and be-
tween the same p!|allels AD, E AD F
BC : the triangle ABC^s equal to
the triangle DBC.

Produce AD both wblfa to the
points E, F, and through B draw
(31. 1.) BE parallel to C\; and
through C draw CP pSjrallel
to BD: therefore each oiythe
figures EBCA, DBCF is a paWlelogram : and EBCA is equpil (35. 1.)
to DBCF, because they are u|^n the base BC, and between the
same parallels BC, EF ; and thf triangle ABC is the half of the pa-
rallelogram EBCA, because theldiameter AB bisects (34. 1.) it; and
the triangle DBC is the half of t^P parallelogram DBCF, because the
diameter DC bisects it : but the balve$ of equal things are equal :
(7. Ax.) therefore the triangle AnC is equal to the triangle DBC.
Wherefore triangles, &c. Q,. E. E^

PROP. XXXFin. THEOR.

Triangles upon equal bases, and between the same parallels^
are equal to one another.

Let the triangles ABC, DEF be upon equal bases BC, EF, and

BOOK I.

TBE SLEMEMTB OP BUCUD.

between the same peurallds BF« AD; the triangle ABC is equal to
the triangle DEF.

Produce AD both ways to the points G, H, and through B draw
BG parallel (31. 1.) to CA, and through F draw FH parallel to ED :
then each of the figures G A « D H

GBCA,D£FH is a paralldo-
gram; and they are equal
(36. 1.) to one another, be-
cause they are upon equal
bases BC, EF, and between
the same parallels BF, GH ;
and the triangle ABC is the
half (34. 1.) of the parallelo-
gram GBC A, because the diameter AB bisects it ; and the triangle
DEF is the half (34. 1.) of the parallelogram DEFH, because the
diameter DF bisects it : but the halves of equal things are equal ;
(7. Ax.) therefore the triangle ABC is equal to the triangle DEF.
Wherefore triangles, &c. Qi. E. D.

PROP. XXXIX. THEOR.

ESqual triangles upon the same base, and upon the same side
of it, are between the same parallels.

Let the equal triangles ABC, DBC be upbn the same base BC,
and upon the same side of it ; they are between the same parallels.

Join AD; AD is parallel to BC : for, if it is not, through the point
A draw (31. 1.) AE parallel to BC, and join EC ; the triangle ABC
is equal (37. 1.) to the triangle EBC, because it is upon the same
base BC, and between the same parallels BC, A D

AE : but the triangle ABC is equal to the tri-
angle BDC ; therefore also the triangle BDC
is equal to the triangle EBC, the greater to the
less, which is impossible ; therefore AE is not
parallel to BC. In the same manner, it can
be demonstrated that no other Ime but AD is
parallel to BC : AD is therefore parallel to it.
Wherefore equal triangles upon, &ic. Q,. E. D.

PROP. XL. THEOR.

Equal triangles upon equal bases, in the same straight line,
and towards the same parts, are between the same parallels.

Let the equal triangles ABC, DEF be upon equal bases BC,
EF, in the same straight line A ~

BF, and towards the same parts;
they are between the same pa-
rallels.

Join AD; AD is parallel to
BC; for if it is not, through A
draw (31. 1.) AG parallel to
BF, and join GFi the triangle B

34 THE ELEMENTS OF EUCLID. BOOK 1.

ABC is equal (38. 1.) to the triangle GEF, because they are upon
equal bases BC, EF, and between the same parallels BF, AG : but
the triangle ABC is equal to the triangle DEF ; therefore also the
triangle DEF is equal to the triangle GEF, the greater to the less,
which is impossible ; therefore AG is not parallel to BF : and in the
same manner it can be demonstrated that there is no other parallel
to it but AD ; AD is therefore parallel to BF. Wherefore, equal tri-
angles, &c. Q. E. D.

PROP. XLL THEOR.

If a paralellogram and triangle be upon the same base, and
between the same parallels ; the parallelogram shall be double
of the triangle.

Let the parallelogram ABCD and the triangle EBC be upon the
same base BC, and between the same parallels BC, AE ; the paral-
lelogram ABCD is double of the triangle EBC. A D E

Join AC ; then the triangle ABC is equal
(37. 1.) to the triangle EBC, because they
are upon the same base BC, and between the
same parallels BC, AE. But the parallelo-
gram ABCD is double (34. 1.) of the trian-
gle ABC, because the diameter AC divides it
into two equal parts; wherefore ABCD is
also double of the triangle EBC. Therefore
if a parallelogram, &c. Q. E. D.

PROP. XLIL PROB.

To describe a parallelogram that shall be equal to a given
triangle, and have one of its angles equal to a given fectilmeal
■ angle.

Let ABC be the given triangle, and D the given rectilineal angle.
It is required to describe a parallelogram that shall be equal to the
given triangle ABC, and have one of its angles equal to D.

Bisect (10. 1.) BC in E, join AE, and at the point E in the
straight line EC make (23. 1.) the angle CEF equal to D ; and
through A draw (31. 1.) AG parallel to EC and through C draw
CG (31. 1.) parallel to EF : therefore A F G

FECG is a parallelogram : and be-
cause BE is equal to EC, the trian-
gle ABE is likewise equal (38. 1.)
to the triangle A EC, since they are
upon equal bases BE, EC and be-
tween the same parallels BC, AG;
therefore the triangle ABC is double
of the triangle AEC ; and the paral-
lelogram FECG is likewise double
(41. 1.) of the triangle AEC, be-
cause it is upon the same base, and between the same parallels :

BOOK I.

THE ELEMI»T9 OP EUCLID.

therefore the paralldogram FECG is equal to the triangle ABC, and
it has CHie of its angles CEF equal to the given angle D. Wherefore
there has been described a parallelogram FECG equal to a given
triangle ABC, having one of its angles CEF equal to the given angle
D. Which was to be done.

PROP. XLHL THEOR.

The complements of the parallelograms i^hich are about the
diameter of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, of which the diameter is AC, and
EH, FG the parallelograms about AC, AH D

that is, through which AC passes, and
BK, KD the other parallelograms which
make up the whole figure ABCD, which E
are therefore called the complements :
the complement BK h equal to the com-
plement KD.

Because ABCD is a parallelogram, i
and AC its diameter, the triangle ABC
is equal (34. 1.) to the triangle ADC; ^
and because EKHA is a parallelogram, the diameter of which is AK,
the triangle AEK is equal to the triangle AHK : by the same reason,
the triangle K<3rC is equal to the triangle KFC : then, because the
triangle AEK is equal to the triangle AHK, and the triangle KGC to
KFC ; the triangle AEK, together with the triangle KGC, is equal to
the triangle AHK together with the triangle KFC: but the whole
triangle ABC is equal to the whole ADC ; therefore the remaining
complement BK is equal to the remaining complement KD. Where-
fore the complements, &c. Q^ E. D.

PROP. XLIV. PROB.

To a given straight line to apply a parallelogram, which shall
be equal to a given triangle, and have one of its angles equal to
a given rectilineal angle.

Let AB be the given straight line, and C the given triangle, and
D the given rectilineal angle. It is required to apply to the straight
line AB a parallelogram equal to the triangle C, and having an angle
equal to D.

Make (42. 1.) the F E K

parallelogram BEFG
equal to the trian-
gle C, and having
the angle EBG equal
to the angle D, so
that BE be in the
same straight line
with AB, and pro-
duce FG to H ; and

i *

36 THE iSLeM&NTS OF CitrcLlD. BOOS 1.

through A draw (31. 1.) AH parallel to BG 6r EP^ and join HB.
Then, because the straight line HF fells upon the parallels AH, £F,
the rfigles AHF, HFE are together equal (29. 1.) to two right angles :
wherefore the angles BHF, HFE are less than two right angles : but
straight lines which with another straight line make thd interior
angles upon the same side less than two right angles, do meet t^(12.
Ax.) if produced far enough : therefore HB, FE shall meeti if pro-
duced ; let them meet in K, and through K, draw KL parallel to EA
or FH, and produce HA, GB to the points LM : then HLKF is a pa-
rallelogram, of which the diameter is HK, and AG, ME are the pa-
rallelograms about HK ; and LB, BF are the complements ; there-
fore LB is equal (43. 1.) to BF ; but BF is equal to the triangle C :
wherefore LB is equal to the triangle C : and because the angle
GBE is equal (15. 1.) to the angle ABM, and likewise to the angle
D ; the angle ABM is equal to the angle D : therefore the parallelo-
gram LB is applied to the straight line, AB, is equal to the triangle
C, and has the angle ABM equal to the angle D. Which was to be
done.

PROP. XLV. PROB.

To describe a parallelogram equal to d. given retilineal figure,
and having an angle equal to a given rectilineal angle.*

Let ABCD be the given rectilineal figure, and E the given recti-
lineal angle. It is required to describe 3, parallelogram equal to
ABCD, and having an angle equal to E.

Join DB, and describe (42. 1.) the parallelogram FH equal to
the triangle ADB, and having the angle HKF equal to the angle
E ; and to the straight line GH apply (44. 1.) the parallelogram
GM equal to the triangle DBC, having the angle GHM equal to
the angle E ; and because the angle E is equal to each of the an-
gles FKH, GHM, the angle FKH is equal to GHM : add to each
of these the angle KHG ; therefore the angles FKH, KHG, are
equal to the angles A D F G L

KHG, GHM ; but
FKH, KHG are equal

(29. 1.) to two right \ ./ \ / ^

angles : therefore also
KHG, GHM are equal
to two right angles ;
and because at the
point H in the straight
Ime GH, the two
straight lines KH, HM, upon the opposite sides of it, make the
adjacent angles equal to two right angles, KH is in the same
straight line (14. 1.) with HM, and because the straight line HO
meets the parallels KM, FG; the alternate angles MHG, HGF
are equal: (29. L) add to each of these the angle HGL; therefore

* See Note.

BOOK I. TRB SLtMBMTS OP lUCLIO. 37

the ang!e9 MHO, BOfL axt equal to the angles HGF, HGL ; but the
angles. HGM, HGL are equal (29. 1.) to two right angles; where*
fore also the angles HGF, HGL are equal to two right angles, and
FG is therefore in the same straight line with GL : and because KF
is parallel to HG, and HG to ML ; KF is parallel (30. L) to ML ; and
KM, FL are parallels ; wherefore EFLM is a parallelogram; and be-
cause the triangle ABD is equal to the parallelogram HF, and the
triangle DBG to the parallelogram GM ; the whole rectilineal figure
ABCD is equal to the whole parallelogram KFLM ; therefore the
parallelogram KFLM has been described equal to the given rectilineal
figure ABCD, having the angle FKM equal to the given angle E.
Which was to be done.

Cor. From this it is manifest how to a given straight line to apply
a parallelogram, which shall have an angle equsd to a given rectilineal
angle, and shcdl be equal to a given rectilineal figure, viz. by apply-
ing (44. 1.) to the given straight line a parallelogram equal to the
first triangle ABD, and having an angle equal to the given angle.

PROP. XLVL PROB.

To describe a square upon a given straight line.

Let AB be the given straight line ; it is required to describe a
square upon AB.

From the point A draw (II. 1.) AC at right angles to AB; and
make (3. 1.) AD equal to AB, and^ through the point D draw DE
parallel (31. 1.) to AB, and through B draw BE parallel to AD;
therefore ADEB is a parallelogram: whence AB is equal (34. 1.) to
DE, and AD to BE ; but BA is equal to AD ; there- q
fore, the four straight lines BA, AD, DE, EB are
equal to one another, and the parallelogram ADEB
is equilateral, likewise all its angles are right an-
gles : because the straight lines AD meeting the
parallels AB, DE, the angles BAD, ADE are equal D
(29. 1.) to two right angles : but BAD is a right
angle; therefore also ADE is a right angle; but
the opposite angles of parsrilelograms are equal,
(34. 1.) therefore each of the opposite angles ABE, ^
BED is a right angle ; wherefore the figure ADEB
is rectangular, and it has been demonstrated that it is equilateral ; it
is therefore a square, and it is described upon the given straight line
AB. Which was to be done.

Cor. Hence every paralieiogram that has one right angle, has all
its angles right angles.

PROP. XLVn. THEOR.

In any right angled triangle, the square which is described

THE GLEMBITTS OF EUCLID.

BOOK I.

Upon the side subtending the right angle, is equal to the squares
described upon the sides which contain the right angle.

Let ABC be a right angled triangle, having the right angle BAG;
the square described upon the side BC is equal to the squares de-
scribed upon BA, AC.

On BC describe (46. 1.) the square BDEC, and on BA, AC the
squares GB, HC: and through A draw (31. 1.) AL parallel to BD or
CE, and join AD, FC : then, because each of the angles BAC, BAG,
is a right angle, (30. def.) the g

two straight Hnes AC, AG upon
the opposite sides of AB, make
with it at the point A the adjacent
angles equal to two right angles : F
therefore CA is in the same . ,. , ...

straight line (14. 1.) with AG:
for the same reason, AB and AH
are in the same straight line ; and
because the angle DBC is equal
to the angle FBA, each of them
being a right angle, add to each
the angle ABC, and the whole an-
gle DBA is equal (2. Ax.) to the
whole FBC : and because the two
sides AB, BD are equal to the two
FB, BC, each to each, and the angle DBA equal to the angle FBC ;
therefore jthe base AD is equal (4. 1.) to the base FC, and the trian-
gle ABD to the triangle FBC : now the parallelogram BL is double
(41. I.) of the triangle ABD, because they are upon the same base
BD, and between the same parallels BD, AL ; and the square GB is
double of the triangle FBC, because these also are upon the same
base FB, and between the same parallels FB, GC. But the doubles
of equals are equal (6. Ax.) to one another : therefore the parallelo-
gram BL is equal to the square GB : and in the same manner, by
joining AE, BE, it is demonstrated that the parallelogram CL is equal
to the square HC : therefore the whole square BDEC is equal to the
two squares GB, HC ; and the square BDEC is described upon the
straight line BC, and the squares GB, HQ upon BA, AC : wherefore
the square upon the side BC is equal jto the squares upon the sides
BA, AC. Therefore, in any right angled triangle, &c. Q. E. D.

PROP. XLVni. THEOR.

If the square described upon one of Ihe sides of a triangle be
equal to the squares described upon the other two sides of it ; the
angle contained by these two sides is a right angle.

If the square described upon BC, one of the sides of the triangle
ABC, be equal to the squares upon the other sides BA, AC, the an-
gle BAC is a right an^e.

BOOK I. THE £LEM£IIT8 OF EUCLID. 39

From Hie point A draw (11. 1.) AD at right angles to AC, and
make AD equal to BA, and join DC : then because DA is equal to
AB, the square of DA is equal to the square of D

AB : to each of these add the square of AC :
therefore the squares of DA, AC are equal to the
squares of BA, AC : but the square of DC is .equal
(47. 1.) to the squares of DA, AC, because DAC
is a right angle ; and the square of BC, by hypO"
thesis, is equal to the squares of BA, AC; there-
fore the square of DC is equal to the square of
BC ; and therefore also the side DC is equal to
the side BC. And because the side DA is equal ^ C

to AB, and AC common to the two triangles DAC, BAC, the two
DA, AC are equal to the two BA, AC : and the base DC Is equal to
the base BC : therefore the angle DAC is equal (8. 1.) to the angle
BAC: but DAC is a right angle ; therefore also BAC is a right angle.
Therefore, if the square, &c. Q- E. D»

THE

ELEMENTS OF EUCLID

BOOK 11.

DEFINITIONS.

EVERY right angled parallelogram is said to be contained by any
two of the straight lines which contain one of the right angles.

In every parallelogram, any of the parallelograms about a diameter,
together with the two complements is E

called a gnomon. * Thus the paral- A

* lelogram HG, together with the com-
^plements AF, FC, is the gnomon,

• which is more briefly expressed by
•the letters AGK, or EHC, which are
•at the opposite angles of the paral- ^
•lelograms which make the gnomon.*

PROP. I. THEOR.

If there be two straight lines, one of which is divided into
any number of parts; the rectangle contained by the two
straight lines, is equal to the rectangles contained by the un-
divided line, and the several parts of the divided line.

Let A and BC be two straight lines ; and let BC be divided into
any parts in the pomts D, E ; the rectangle contained by the straight
lines A, BC is equal to the rectangle B DEC

contained by A, BD, together with that
contained by A, DE, and that contained
by A, EC.

From the point B draw (II. 1.)
BF at right angles to BC, and make
BG equal (3. I.) to A; and through q
G draw (31. 1.) GH paraDel to BC; and
through D, Ei C draw (31. 1.) DK, .

EL, CH paraUel to BG: .then the I ^

rectangle BH is equal to the rect- F
angles BK, DL, EH ; and BH is contained by A, BC, for it is

B^OOK II.

Tlifi XLBMBirrfl OF EUCLID.

€1

eoQtained by OB, BC, and GB is equal to A ; and BK is contained
by A, BD, fot it is contained by QB, BD, of which GB is eqaal to A;
and DL is contained by A, DE, because DK, that is (34. 1.) BG is
equal to A ; and in like manner the rectangle £H is contained by A,
EC : therefore the rectangle contained by A, BC is equal to the several
rectangles contained by A, BD, and by A, DE; and also by A, EC.
Wherefore, if there be two straight lines, &c. Q. E. D.

PROP. n. THEOR.

If a straight line be divided into any two parts, the rectangles
contained by the whole and each of the parts, are together equal
to the square of the whole line.

Let the straight line AB be divided into any A C B

two parts in the point C; the rectangle contained
by AB, BC, together with the rectangle* AB, AC,
shall be equal to the square of AB.

Upon AB describe (46. 1.) the square ADEB,
and through C draw (31. 1) CF, parallel to AD
or BE ; then AE is equal to the rectangles AF,
CE : and AE is the square of AB : and AF is the
rectangle contained by BA, AC ; for it is contain-
ed by DA, AC, of which AD is equal to AB; and D F _
Cfi is contained by AB, BC, for BE is equal to AB ; therefore the
rectangle contained by AB, AC, together with the rectangle AB, BC,
is equal to the square of AB. If therefore a straight line, &c, Q^ E. D.

PROP. m. THEOR.

If a straight line be dtyided into any two parts, the rectangle
contained by the whole and one of the parts, is equal to the rect-
angle contained by the two parts, together with the square of the
foresaid part

L^t the straight line AB be divided into two parts in the point C ;
the rectangle AB, BC is equal to the rect^gle AC, GB togeth^ with
the square of BC.

Upon BC describe (46. 1.) the square AC B

'CDEB, and produce ED to F, and
through A draw (31. 1.) AF parallel to
CD or BE; then the rectangle AE is
equal' to the rectangles AD, CE; and
AE is the rectangle contained by AB,
BC, for it is contained by AB, BE, of
which BE is equal to BC; and AD is
contained by AC, CB, for CD is equal
to BC; and DB is the squat^e of BC; F

■ • ■

^1 ■ Jill ■»W^i— fc^a^p— ^— ^■^W — ^"^

* N. B. To avoid repeating the word tontained too frequently, the rectangle con-
tained by two straight lines AB, AC is sometimes simply called the rectangle AB,
AC.

THB BLGMBKTS OF BUCLFD.

BOOK II.

therefore the rectangle AB, BC* is equal to the rectangle AC, GB, to-
gether with the square of BC. If therefore a straight line, &c. ^ E. D.

PROP. IV. THEOR.

Ir a straight line be divided into any two parts, the square of
the whole line is equal to the squares of the two parts, together
with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C ; the
square of AB is equal to the squares of AC, CB, and to twice the
rectangle contained by AC, CB.

Upon AB describe (46. 1.) the square ADE6, and join BD, and
through C draw (31. I.) CGF parallel to AD or BE and through G
draw HK parallel to AB or DE : and because CF is parallel to AD,
and BD falls upon them, the exterior angle BGC is equal (29. I.) to
the interior and opposite angle ADB ; but ADB is equal (5. 1.) to the
angle ABD, because BA is equal to AD, being sides of a square ;
wherefore the angle CGB is equal to the A C B

angle GBC ; and therefore the side BC is
equal (6. 1.) to the side CG; but CB is
equal (34. 1.) also to GK, and CG to BK;
wherefore the figure CGKB is equilateral ;
it is likewise rectangular ; for CG is paral-
M to BK, and CB meets them; the angles
RBC, GCB are therefore equal to two right
angles ; and KBC is a right angle ; where-
fore GCB is a right angle: and therefore D F
also the angles (34. 1.) CGK, GKB, opposite to these, are right angles,
and CGKB is rectangular : but it is also equilateral, as was demon-
strated ; wherefore it is a square, and it is upon the side CB : for the
same reason HF also is a square, and it is upon the side HG, which
is equal to AC : therefore HF, CK are the squares of AC, CB ; and
because the complement AG is equal (43. 1.) to the complement GK^
and that AG is the rectangle contained by AC, CB, for GC is equal
to CB; therefore GE is also equal to the rectangle AC, CB; where-
fore AG GE are equal to twice the rectangle AC, CB; and HF, CK
are the squares of AC, CB : wherefore the four figures HF, CK, AG,
GE are equal to the squares of AC, CB, and to twice the rectangle
AC, CB, but HF, CK, AG, GE make up the whole figure ADEB,
which is the square of AB : therefore the square of AB is equal to the
squares of AC, CB, and twice the rectangle AC, CB. Wherefore,
if a straight line, &;c. Q. E. D.

Cor. From the demonstration it is manifest, that the parallelograms
about the diameter of a square are likewise squares.

PROP. V. THEOR.

If a straight line be divided into two equal parts, and also into

BOOK II.

THE BL£lf EMTS OF SUCUD.

two unequal parts: the rectangle contained by the unequal parts^
together ^ith the square of the line between the points of section^
is equal to the square of half the line.

Let the straight line A6 be divided into two equal parts in the
point C, and into two unequal parts at the point D ; the rectangle
AD, DB, together with the square of CD, is equal to the square of
CR

Upon CB describe (46. 1.) the square CEFB, join BE, ancJ
through D draw (31. 1.) DHG parallel to CE or BF; and through
H draw KLM parallel to CB or EF ; and also through A draw
AK parallel to CL or BM ; and because the complement CH is
equal (43. 1.) to the complement HF, to each of these add DM;
therefore the whole CM is equal AC D B

to the whole DF; but CM is
equal (36. 1.) to AL, because AC
is equal to CB; therefore also
AL is equal to DF. To each of
tiiese add CH, and the whole
AH is equal to DF and CH:
but AH is the rectangle con-
tained by AD, DB) for DH is
equal (Cor. 4. 2.) to DB, and
DF together with CH is the gnomon CMG; therefore the gno-
mon CMG is equal to the rectangle AD, DB: to each of these
add LG, which is equal (43. 1.) to the square of CD; therefore
the gnomon CMG, together wit^i LG, is equal to the rectangle AD,
DB, together with the square of CD : but the gnomon CMG and LG
makes up the whole figure CEFB, which is the square of CB: there*
fore the rectangle AD, DB, together with the square of CD, is equal to
the square of CB. Wherefore, if a straight line, &c. d. E. D.

From this proposition it is manifest, that the difference of the
squares of two unequal lines, AC, CD, is equal to the rectangle con-
tained by their sum and difference.

PROP. VL THEOR*

If a straight line be bisected, and produced to any jpoint ; the
rectangle contained by the whole line thus produced, and the
part of it produced, together with the square of half the line bi-
sected, is equal to the square of the straight fine which is made
up of the half and the part produced.

Let the straight line AB be bisected in C, and produced to the
point D ; the rectangle AD, DB, together with the square of CB, i»
equal to the square of CD.

Upon CD describe (46. 1.) the square of CBFD, join DE, and
through B draw (31. 1.) BHG paraUd to CE or DF, and througk

THB ELfiBIEll^S OF EUCLID.

Booit n.

G P

H draw KLM parallel to AD or £F, and also through A draw AK
parallel to CL or DM : wd be- A C B D

cause AC is equal to CB, the rec-
tangle AL is equal (43. 1.) to CH;
but CH is equal (36. 1.) to HF ; K
therefore also AL is equal to HF:
to each of these add CM ; there- ,
fore the whole AM is equal to the
gnomon CMG: and DM is the
rectangle contained by AD, DB, fpr DM is equal (Cor. 4. 3.) to DB :
therefore the gnomon CMG is equal to the rectangle AD, DB, add
to each of these LG, which is equal to the square of CB, therefore
the rectangle AD, DB, together with the square of CB, is equal to
the gnomon CMG and the figure LG : but the gnomon CMG apd
LG make up the whole figure CEFD, which is the square of CD ;
therefore the rectangle AD, DB, together with the square of CB, is
equal to the square of CD. Wherefore, if a straight line, &c.
d. £. D.

PROP. VII. THEOR.

If a straight line be divided into any two parts, the squares of
the whole line, and of one of the parts are equal to twice the rect-
angle contained by the whole and that part, together with the
square of the other part.

Let the strai^t line AB be divided into any two parts in the point
G ; the squares of AB, BC are equal to twice the rectangle AB, BC,
together with the square of AC.

Upon AB desmbe (46. 1.) the square ADEB, and construct the
figure as in the preceding propositions : and because AG is equal
(43. 1.) to GE, add to each of them CK; the whole AK is therefore
equal to the whole CE; therefore AK, CE A C B

are double of AK : but AK, CE are the
gnomon AKF, together with the square
CK ; therefore the gnomon AKF, together
with the square CK, is double of AK : but H
twice the rectangle AB, BC is double of AK,
for BK is equal (Cor. 4. 2.) to BC ; there*
fore the gnomon AKF, together with the
square CK, is equal to twice the rectangle
AB, BC : to each of these equals add HF, D F E

which is equal to the square of AC : therefore the gnomon AKF,
together with the squares CK, Hfe', is equal to twice the rectangle
AB, BC, and the square of AC: but the gnomon AKF, together
with the squares OK, HF, make up the whole figure ADEB and
CK, which are the squares of AB and BC, therefore the squares of
AB and BC are equal to twice the rectangle AB, BC, together with
the square AC. Wherefore, if a straight line, &c. Q. E. D.

B0(» II.

TH£ BLEMENTfl Or EUCLID.

PROP. Vm. THEOR.

If a straight line be divided into any two parts, four times the
rectangle contained by the whole line, and one of the parts, to-
gether with the square of the other part, is equal to the square
of the straight line which is made up of the whole and that part

•

Let the straight line AB be divided into any two parts in the point
C ; four times the rectangle AB, BC, together with the square of AC,
is equal to the square of the straight line made up of AB and BC
together.

Produce AB to D, so that BD be equal to CB, and upon AD de*
scribe the square AEFD ; and construct two figures such as in the
preceding. Because CB is equal to BD, and that CB is equal
(34. 1.) to GK, and BD, to KN; therefore GK is equal to KN;
for the same reason, PR is equal to RO ; and because CB is equal
to BD, and GK to KN, the rectangle CK is equal (36. 1.) to BN,
and GR to RN ; but CK is equal (43. 1.) to RN, because they are
the complements of the parallelogram CO; therefore also BN is
equal to GR; and the four rectangles BN, CK, GR, RN are
therefore equal to one another, and so are quadruple of one rf
them CK: again, because CB is equal to BD, and that BD is
equal (Cor. 4. 2.) to BK, that is, to CG ; C B

and CB equal to GK, that (Cor. 4. 2,) A
is, to GP ; therefore CG is equal to GP :
and because (Xr is equal to GP, and PR M ^
to RO, the rectangle AG is equal to
MP, and PL to RF : but MP is equal X
(43. L) to PL, because they are the
complements of the parallelogram ML;
wherefore AG is equal also to RF:
therefore the four rectangles AG, MP,
PL, RF are equal to one another, and so
are quadruple of one of them AG. And it was demonstrated that
the four CK, BN, GR, and RN are quadruple to CK: therefore the
eight rectangles which contain the gnomon AOH are quadruple of
AK : and becaluse AK is the rectangle contained by AB, BC, for BK
is equal to BC, four times the rectangle AB, BC is quadruple of AK:
but the gnomon AOH was demonstrated to be quadruple of AK : there'
fore four times the rectangle AB, BC is equal to the gnomon AOH.
To each of these add XH, which is equal (Cor. 4. 2.) to the square
of AC : therefore four times the rectangle AB, BC, together with the
square of AC, is equal to the gnomon AOH and the square XH : but
the gnomon AOH and XH make up the figure AEFD, which is the
square of AD : therefore four times the rectangle AB, BC, together
with the square of AC is equal to the square of AD, that is, of AB
and BC added together hi one straight line. Wherefore, if a straight
line, &c. O. £. D.

40 THE BLBMENTS OP EUCLID. BOOK U»

PROP. IX. THEOR.

If a straight line be divided into two equal, and also into two
unequal parts; the squares of the two unequal parts are to-
gether double of the square of half the line, and of the square of
the line between the points of section.

Let the straight lihe AB be divided at the point C into two equal,
and at D into two unequal parts : the squares of AD, DB are to-
gether double of the squares of AC, CD.

From the point C draw (11. 1.) CE at right angles to AB, and
mAe it equal to AC or CB, and join EA, EB ; through D draw
(31. 1.) DF parallel to CE, and through F draw FG parallel to AB ^
and join AF ; then, because AC is equal to CE, the angle EAC is
equal (5. 1.) to the angle AEC; and because the angle ACE is a right
angle, the two others AEC, EAC together make one right angle
(32. 1.); and they are equal to one another ; each of them therefore
is half of a right angle. For the same A

reason each of the angles CEB, EBC is
half a right angle; and therefore the
whole AEB is a right angle : and because
the angle GEF is half a right angle, and
BGF a right angle, for it is equal (29. 1.)
to the interior and opposite angle ECB,
the remaining angle EFG is half a right ^
angle ; therefore the angle GEF is equal to the angle EFG, and the
side EG equal (6. 1.) to the side GF : again, because the angle at B
is half a right angle, and FDB half a right angle, for it is equal (29.
1.) to the interior and opposite angle ECB, the remaining angle BFI>
is half a right angle ; therefore the angle at B is equal to the angle
BFD, and the side DF to (6. 1.) the side DB : and because AC is
equal to CE, the square of AC is equal to the square of CE ; there-
fore the squares of AC, CE are double of the square of AC : but
the square of EA is equal (47. 1.) to the squares of AC, CE, because
ACE is a right angle ; therefore the square of EA is double of the
square of AC : again, because EG is equal to GF, the square of EG is
equal to the square of GF ; therefore the squares of FG, GF are double
of the square of GF ; but the square of EF is equal to the squares
of EG, GF ; therefore the square of EF is double of the square of
GF; and GF is equal (34. I.) to CD ; therefore the square of EF is
double of the square of CD: but the square of AE is likewise douWe
of the square of AC : therefore the squares of AE, EF are double
of the squares of AC, CD ; and the square of AF is equal (47. 1.) io
the squares of AE, EF, because AEF is a right angle ; therefore the
square of AF is double of the squares of AC, CD : but the squares
of AD, DF are equal to the square of AF, because the angle ADF
is a right angle ; therefore the squares of AD, DF are doubje of the
squares of AC, CD: and DF is equal to DB; therefore the squares
of AD, DA are double of the squares of AC, CD. If therefore &
straight line, &c. CI. E. D.

BOOK. fl. THB ELEMENTS OF EUCLID. 47

PROP. X. THEOR.

If a straight line be bisected, and produced to any point, the
square of the whole line thus produced and the square of the
part of it produced, are together double of the square of half
the line bisected, and of the square of the line made up of the
half and the part produced.

Let the straight line AB be bisected in C, and produced to the
point D ; the squares of AD, DB are double of the squares of AC,
CD.

From the point C draw (11. 1.) C£ at right angles to AB: and
make it equal to AC, or CB, and join AE, EB ; through £ draw
(31. 1.) EF parallel to AB, and through D draw DF, parallel to CE:
and because the straight line EF meets the parallels EC, FD, the
angles CEF, EFD are equal (29. 1.) to two right angles ; and there-
fore the angles BEF, EFD are less than two right angles: but
straight lines which with another straight line make the interior
angles upon the same side less than two right angles do meet (12.
Ax.) if produced far enough : therefore EB, FD shall meet, if pro-
duced towards B, D: let them meet in G, and join AG: then,
because AC is equal to CE, the angle CEA is equal (5. 1.) to the
angle EAC : and the angle ACE is a right angle : therefore each of
the angles CEA, EAC is half a right angle (32. 1.) : for the same
reason, each of the angles CEB, EBC is half a right angle : there-
fore AEB is a right angle : and because EBC is half a right angle,
DBG is also (15. 1.) half a right angle, for they are vertically oppo-
site; but BDG is a right angle, because it is equal (29. 1.) to the
alternate angle DCE ; therefore the remaining angle DGB is half a
right angle, and is therefore equal to the angle DBG; wherefore
also the side BD is equal (6. 1.) to the side DG: again, because
EGF is half a right angle, and that the angle at F is a right angle,
because it is equal (34. 1 .) to the E F

opposite angle ECD, the remain-
ing angle FEG is half a right
an^e, and equal to the angle
EGF; wherefore also the side
OF is equal (fi. l.) to the side / ^

F£. And because EC is equal A*
to C A, the square of EC is equal
t9 the square of CA ; therefore
tlie squares of EC, CA are doti- G

ble of the square of CA : but the square of EA is equal (47. 1.) to
the squares of EC, C A ; therefore the square of EA is double of the
square of AC : again, because GF is equal to FE, the square of
GF is equal to the square of FE: and therefore the squares of
GF, FE are doubfe of the square of EF : but the square of EG is
equal (47. 1.) to the squares of GF, FE; therefore the square of

THK fiLEMENTS OP WCUO,

EOOK ii«

EG Is double of the square of EF : and EF is equal to CD; where^
fore the square of EG is double of the square of CD : but it was de-
monstrated, that the square of EA Is double of the square of AC ;
therefore the squares of AE, EG are double of the squares of AC,
CD : and the square of AG is equal (47. 1.) to the squares of AE,
EG ; therefore the square of AG is double of the squares of AC, CD :
but the squares of AD, QD are equal (47. 1.) to the square of AG ;
therefore the squares of AD, DG are double of the squares of AC,
CD : but DG is eqtial to DB ; therefore the squares of AD, DB are
double of the squares of AC, CD. Wherefbi'e, if a straight line, &c.
Q. E. D.

PROP. XI. PROB.

To divide a given straight line into two parts, so that the rect-
angle contained by the whole and one of the parts shall be
equal to the square of the other part

Let AB be the given straight line : it is required to divide ft into
two parts, so that the rectangle contained. by the whole and one of
the parts shall be equal to the squai^e of the other part

Upon AB describe (46. 1.) the square ABDC ; bisect (10. 1«) AC
in E, and join BE; produce CA to F, and make (3^ 1.) EF equal
to EB; and upon AF describe (46. 1.) the 'square FGHA; AB is
divided in H, so that the rectangle AB, BH is equal to the square
of AH.

Produce GH to K ; because the straight line AC is bisected in E,
and produced to the point F, the rectangle CF, FA, together with
the square of AE, is equal (6. *^.) to the square of EF : but EF is
equal to EB; therefore the rectangle CF, FA, together with the
square of AE, is equal to the square of EB : and the squares of
BA, AE are equal (47. 1.) to the square of F G

EB, because the angle EIAB is a right angle ;
therefore the rectangle CF, FA, together
with the square of AE, is equal to the squares
of BA, AE : take away the square of AE,
which is common to both, therefore the re-
maining rectangle CF, FA is equal to the A""
square of AB ; and the figure FK is the rec-
tangle contained by CF, FA, for AF is equal
to FG ; and AD is the square of AB ; there-
fore FK is equal to AD : take away the E
common part AK, and the remainder FH is
equal to the remainder HD, and HD is the
rectangle contained by AB, BH, for AB it
equal to BD ; and FH is the square of AH : C K D

therefore the rectangle AB, BH is equal to the square of AH : where-
fore the straight line AB is divided in H, so that the rectangle AB,
BH is equal to the square of AH. Which was to be done.

BOOK II. THB SLBMIMT9 OP EUCLID. 4Q

PROP. Xn. THEOR.

Iir obtuse angled triangles, if a perpendicular be drawn from
any of the acute angles to the opposite side produced, the square
of the side subtending the obtuse angle is greater than the squares
of the sides containing the obtuse angle, by twice the rectangle
contained by the side upon which, when produced, the perpen-
dicular falls, and the straight line intercepted without the trian-
gle between the perpendicular and the obtuse angle.

Let ABC be an obtuse angled triangle, having the obtuse angle
ACB, and from the point A let AD be drawn (12. 1,) perpendicular
to BC produced : the square of AB is greater than the squares of
AC CB, by twice the rectangle BC, CD.

Because the straight line BD is divided into two parts in the point
C, the square of BD is equal (4. 2.) to
the squares of BC, CD, and twice
the rectangle BC, CD: to each of
these equals add the square of DA ;
and the squares of BD DA are equal
to the squares of BC, CD, DA, and
twice the rectangle BC, CD : but the
square of BA is equal (47. 1.) to the
squares of BB, DA, because the angle
at D is a right angle; and the square
of CA is equal (47. 1.) to the squares
of CD, DA : therefore the square of BA is equal to the squares of BO,
CA, and twice the rectangle BC, CD ; that is, the square of BA is
greater than the squares of BC, CA, by twice the rectangle BC, CD.
Therefore, in obtuse angled triangles, &c. Q. E. D.

PROP. Xm. THEOR.

Iir every triangle, the square of the side subtending any of
the acute angles is less than the sl}uares of the sides containing
that angle, by twice the rectangle contained by either of these
sides, and the straight line intercepted between the perpendicu-
lar let fall upon it irom the opposite angle and the acute augle.^

Let BBC be any triangle, and the angle at B one of its acute
angles, and upon BC one of the sides containing it, let fall the per-
pendicular (12. 1.) AD from the opposite angle: the square of AC,
opposite to the angle B, is less than the squares of CB, BA, by twice
the rectangle CB, BD.

* See Note..

THfi BLBMBMTfl OP BUCLID. BOOK II.

First, let AD &11 within the triangle ABC; and because the
straight line CD is divided into two parts in A

the point D, the squares of CB, BD are equal
(7. 2.) to twice the rectangle contained by
CB, BD, and the square of CC : to each of
these equals add the square of AD ; there-
fore the squares of CB, BD, DA are equal to
twice the rectangle CB, BD, and the squares
of AD, DC : but the square of AB is equal /
(47. 1.) to the squares of BD, DA, because ^
the angle BDA is a right angle, and the square B D C

of AC is equal to the squares of AD, DC : therefore the squares of
CB, BA are equal to the square of AC, and twice the rectangle CB,
BD, that is, the square of AC alone is less than the squares of CB,
BA, by twice the rectangle CB, Bfi. A

Secondly, let AD fall without the triangle
ABC : then, because the angle at D is a
right angle, the angle ACB is greater (16.
1.) than a right angle; and therefore the
square of AB is equal (12. 2.) to the squares
of AC, CB, and twice the rectangle BC,
CD : to these equals add the square of BC,
and the squares of AB, BC are equal to the
square of AC, and twee the square of BC, BCD
and twice the rectangle BC, CD: but be-
cause BD is divided into two parts in ; the rectangle DQ, BC is
equal (3. 2.) to the rectangle BC, CD and the square of BC : and
the doubles^of these are equal : therefore the squares of AB, BC are
equal to the squares of AC, and twice the rectangle DB, BC : there-
fore the square of AC alone is less than the square of AB, BC hy
twice the rectangle DB, BC. A

Lastly, let the side AC be perpendicular to BC ; then
is BC the straight line between the perpendicular and
the acute angle at B ; and it is manifest that the square
of AB, BC are equal (47. 1.) to the square of AC and
twice the square of BC. Therefore, in every triangle,
^. Q. K D.

PROP. XIV. PROB.

To describe a square that shall be equal to a given rectilineal
figure.*

Let A be the given rectilineal figure ; it is required to describe
a square that shsdl be equal to A.

Describe (45. L) the rectangular parallelogram BCDE, equal to
the rectilineal figure A. If then the sides of it BE, ED are

* See Note.

BOOK II.

THE BLBUUfTS OF SDCLID.

equal to one another,
it is a square, and
what was required is
now done : but if they
are not equal, pro-
duce one of them BE
to F, and make EP
equal to ED, and bi-
sect BP in G; and
from the centre G, at

the distance GB, or GF, describe the semicircle BHF, and produce
DE to H, and join GH ; therefore because the straight line BF is
divided into two equal parts in the point G, and into two unequal at
E, the rectangle BE, EF, together with the square of EG, is equal (5.
2.) to the square of GF : but GF is equal to GH ; therefore the rect-
angle BE, EF, together with the square of EG, is equal to the
square of GH ; but the squares of HE, EG are equal (47. 1.) to the
square of GH ; therefore the rectangle BE, EF together with the
square of EG, is equal to the squares of HE, EG, : to take away
the square of EG, which is common to both ; and the remaining
rectangle BE, EF is equal to the square of EH : but the rectangle
contained by BE, EF is the parallelogram BD, because EF is equal
to ED; therefore BD is equal to the square of EH; but BD is equal
to the rectOineal figure A ; therefore the rectilineal figure A is equal
to the square of EH: wherefore a square has been made equal to
the given rectilineal figure A, viz. the square described upon EH.
Which was to be done.

THi;

ELEMENTS OF EUCLID

BOOK III.

DEFINITIONS.

Equal circles are those of which the diameters are equal, or from
the centres of which the straight lines to the circumferences are
equal. t

* This is not a definition, but a theorem, the truth of which is evi-
dent ; for, if the circles be applied to one another, so that their cen-
tres coincide, the circles must likewise coincide, since the straight
lines from the centre are equal.'

A straight line is said to touch a
circle, when it meets the cirde,
and being produced does not
cut it.

ra.

Circles are said to touch one ano-
ther, which meet, but do not cut
one another.

IV.

Straight lines are said to be equally distant
from the centre of a circle, when the per-
pendiculars drawn to them from the cen-
tre are equal.

And the straight line on which the greater
perpendicular fells, is said to be further
from the centre.

VI.

A segment of a circle is the figure con-
tained by a straight line and the clr-
cum&rcBce it cuts oflU

BOOK lU.

THS BLBMEIiTS OP BUOUD-

vn.

" The an^e of a segment is that which is contained by the straight
line and the circumference.''

vm.

An angle in a segment is the angle con*
tained by two straight lines drawn from
any point in the circumference of the
segment to the extremities of the
straight line which is the base of the
segment.

IX.

And an angle is said to insist or stand
upon the circumference intercepted be-
tween the straight lines that contain
the angle.

Tiie sector of a circle is the figure con-
tained by two straight lines drawn from
the centre, and the circumference be-
tween them.

XI.

Similar segments of a circle, are
those in which the angles are
equal, or which contain equal
angles.

PROP. I. PROB.
To find the centre of a given circle.*

Let ABC be the given circle; it is required to find its centre.

Draw within it any straight line AB, and bisect (10. 1.) it in D;
from the point D draw (11. 1.) DC at right angles to AB, and pro<-
duce it to £, and bisect CE in F : the point F is the centre of the
circle ABC.

., For, if it be not, let, if possible^ G be the centre, and join G A,
GD, GB : then, because DA is equal to DB, and DG common to
the two triangles ADG, BDG, the two sides AD, DG are equal
to the two BD, DG^ each to each; and the baseGA is equal to
the base GB, because they are drawn from the centre G :t there-
fore the angle ADG is equal (8. 1.) to the angle GDB: but when
a straight line standing upon another straight line makes the ad-

• Sec Note. • ''-*

t N. B. Whenever the ezpreaiion ** straight lines from the centre,*' or " drawn
from the centre,** occurs, it is to be uilderstood that they are dsawn ta the cif-
CDOnftreiice.

THB ILBMBMTS OP BUCUD.

BOOK ni.

jacent angles equal to one another, each of
the aogl^ is a right angle: (10. de£ 1.)
therefore the angle GDB is a right angle ;
but FDB is likewise a right angle ; where-
fore the angle FDB is equal to the angle
GDB, the greater to the less, which is im-
possible : therefore G is iK>t the centre of
the circle ABC: in the same manner it
can be shown, that no other point biat F is
the centre; that is, F is the centre of the
circle ABC. Which was to be found.

CoR« From this it is manifest, that if in
a circle a straight line bisect another at right an
of the circle is in the line which bisects the other.

PROP. n. THEOR.

If any two points be taken in the circumference of a circle,
the straight line which joins them shall fall within the circle.

Let ABC be a circle, and A, B any two points in the circumfe-
rence; the straight line drawn from A to C
B shall fall within the circle.

For, if it do not, let it &11, if possible,
without, as AEB; find (1. 3.) D the centre
of the circle ABC, and join AD, DB, and
produce DF, any straight line meeting the
circumference AB, to E: then because DA
is equal to DB, the angle DAB Is equal
(5. 1.) to the angle DBA ; and because AE,
a side of the triangle DAE, is produced to
B, the angle DEB is greater (16. 1.) than
the angle DAE : but DAE is equal to the angle DBE : therefore the
angle DEB is greater than the angle DBE: but to the greater angle
the greater side is opposite (19. 1.); DB is therefore greater than
D£: but DB is equal to DF; wherefore DF Is greats than DE,
the less than the greater, wbfch is imposs^e : therefore the straight
line drawn from A to B does not M without the circle* &i the
same manner it may be demonstrated that it does not fall upon the
circamforoice ; it Ms therefore within it. Wherefore, if any two
poUits, (Sac. Q. E. D.

PROP. m. THEOR.

>If a straight line drawn through the centre of a circle bisect
a straight hne in it which does not pass through the centre, it
shall cut it at right angles, and» if it cuts it at right angles» it
shall bisect it. '

liCt ABC be a circle ; and let CD, a straight line drawn tiiroHgh

BOOK III.

THE JDLSMBNTS OP EUCLID.

the centre, bisect any straight line AB, which does not pass through
the centre, in the point F : it cuts it also at right angles.

Take (1. 3.) E the centre of the circle, and join EA, EB. Then,
because AF is equal to FB, and FE common to the two triangles
AFE, BFE, there are two sides in the one equal to two sides in the
other, and the base EA is equal to the base C

EB: therefore the angle AFE is equal (8. I.)
to the angle BFE: but when a straight line
standing upon another makes the adjacent
an^es equal to one another, each of them is
a right {10 def.) angle: therefore each of the
angles AFE, BFE is a right angle ; wherefore
the straight line CD, drawn through the cen-
tre bisecting another AB that does not pass j^
through the centre, cuts the same at right
angles*

Let CD cut AB at right angles : CD also bisects it, that is, AF is
equal to FB.

The ssaoae construction being made, because EA, EB from the
centre are equal to one another, the angle EAF is equal (5. 1.) to
the angle EBF : and the right angle AFE is equal to the right angle
BFE: therefore, in the two triangles, EAF, EBF, there are two
angles in one equal to two angles in the other, and the side EF,
which is opposite to one of the equal angles, in each, is common to
both; therefore the other sides are equal (26. 1.): AF therefore is
equal to FB. Whe|:efore, if a straight line, &c. Q. E. D.

PROP. IV. THEOR.

If in a circle two straight lines cut one another which do not
both pass through the centre, they do not bisect each other.

Let ABCD be a circle, and AC, BD two straight lines in it which
cut one another in the point E, and do not both pass through the
centre : AC, BD do not bisect one another.

For, if it is possible, let AE be equal to EC, and BE to ED; if
one of the lines pass through the centre, it is plain that it cannot
be bisected by the other which does not
pass through the centre ; but, if neither
of them pass through the centre, take, (I.
3.) F the centre of the circle, and join
£F: and because FE, a straight line
through the centre, bisects another AC A
which does not pass through the centre,
it shall cut it at right (3. 3.) angles;
wherefore FEA is a right angle: again, c

because the straight tne FE bisects

the straight line BD which does not pass through the centre,
it shall cut it at right <3. 3.) angles; wherefore FEB is a right
angle, and FEA was shown to l^ a right angle ; therefore FEA

X /

THE BLEBISNTt Or EQCblD.

BOOK III*

is equal to the angle FEB, the less to the greater, which is impossi-
ble; therefore AC, BD do not bisect one another. Wherefore, if in a
circle, &c. Q. K. D.

PROP. V. THEOR.

If two circles cut one another, they shall not have the same
centre.

Let the two circles ABC, CDG cut one another in the points B, C;
they have not the same centre.

For, if it be possible, let E be their centre: join EC, and draw
any straight line EFG meeting them in C

F and G ; and because E is the centre
of the circle ABC, CE is equal to EF :
again, because E is the centre of the

circle CDG, CE is equal to EG : but / / \ \ \ G

CE was shown to be equal to EF; f
therefore EF is equal to EG, the less A \
to the greater, which is impossible : \
therefore E is not the centre of the cir- \\,
cles ABC, CDG. Wherefore, if two ^-

circles, &c. Q.. E. D. B

PROP. VI. THEOR.

If tv^o circles touch one another internally, they shall not have
the same centre.

Let the two circles ABC, CDE touch one another internally in the
point C : they have not the same centre.

For, if they can, let it be F ; join FC, and draw any straight line
FEB meeting them in E and B; and C

because F is the centre of the circle ABC,
CF is e^ual to FB; also, because F is
the centre of the circle CDE, CF is equal
to FB: and CF was shown equal to FB;
therefore FE is equal to FB, the less to .
the greater, which is impossible: where-
fore F is not the centre of the circles
ABC, CDE. Therefore, if two circles,
&c. Q. E. D.

PROP. Vn. THEOR.

If any point be taken in the diameter of a circle, v^hich is not
the centre, of all the straight lines which can be drawn from it
to the circumference, the greatest is that in which the centre is,
and the other part of that diameter is the least ; and, of any

BOOK III., TfiS BLBMEHTS OF ECCUD. 57

Others, that which is nearer to the line which passes through the
centre is always greater than one more remote; and from the
same point there can be drawn only two straight lines that are
equal to one another,^ one upon each side of the shortest line.

Let ABCB be a circle, and AD its diameter, in which let any point
F be taken which is not the centre; let the centre be E ; of all the
straight lines FB, FC, FG, &c. that can be drawn from F to the
circumference, FA is the greatest, and FD, the other part of the
diameter AD, is the least : and of the other, FB is greater than FC,
and FC than FG.

Join BE, CE, GE; and because two sides of a triangle are greater
(20. 1.) than the third, BE, EF are greater than BF ; but AE is equal
to EB; therefore AE, EF, that is, AF, is
greater than BF : again, because BE is equal
to CE, and FE common to the triangles
BEF, CEF, the two sides BE, EF are equal S(
to the two CE, EF ; but the angle BEF is
greater than the angle CEF ; therefore the
base BF is greater (24. 1.) than the base
FG: for the same reason OF, is greater
than GF: again, because GF, FE, are greater
(20. 1.) than EG, and EG is equal to ED ;
GF, FE are greater than ED : take away the
common part FE, and the remainder GfF is
greater than the remainder FD : therefore FA is the greatest, and
FD the least of all the straight lines from F to the circumference ;
and BF is greater than CF, and CP than GF.

Also there can be drawn only two equal straight lines from the
point F to the circumference, one upon each side of the shortest line
FD : at the point E, in the straight line EF, make {23. i.) the angle
FEH equal to the angle GEF, and join FH : then because GE is
equal to EH, and EF common to the two triangles GEF, HEF ; the
two sides GE, EF are equal to the two HE EF ; and the angle GEF
is equal to the angle HEF ; therefore the base FG is equal (4. 1.) to
the base FH : but, besides FH, no other straight line can be drawn
from P to the circumference equal to FG, for, if there can, let it be
FK ; and because FK is equal to FG, and FG to FH, FK is equal
to Ell: that is, a line nearer to that which passes through the centre,
is equal to one which is more remote : which is impossible. There-
fore, if any point be taken, &c. Q. E. D.

PROP. Vffl. THEOR.

Ip any point be taken without a circle, and straight lines be
drawn from it to the circumference, whereof one passes through
the centre, of those which fall upon the concave pircumference,
the greatest is that which passes through the centre, and, of the

THE ELEMSirrS OP EUCLID.

BOOK 111.

rest, that which is nearer to that through the centre is always
greater than the more remote : but of those which fall upon the
convex circumference, the ]east is that between the point with-
out the circle and the diameter; and, of the rest, that which is
nearer to the least is always less than the more remote: and only
two equal straight lines can be drawn from the point into the
circumference, one upoQ each side of the least.

Let ABC be a circle, and D any point without it, from which ]et
the straight lines DA, DE, DF, DC be drawn to the circumference,
whereof DA passes through the centre. Of those which fall upon
the concave part of the circumference AEFC, the greatest is AD,
which passes through the centre ; and the nearer to it is always
greater than the more remote, viz. DE than DF, and DF than DC ;
but of those which fall upon the convex circumference HLKG, the
least is DG between the point D a^d; the diameter AG ; and the
nearer to it is always less than the itk)re remote, viz. DK than DL,
and DL than DH.

Take (1. 3.) M the centre of the circle ABC, and join ME, MP,
MC, MK, ML, MH : and because AM is equal to ME, add MD to
each, therefore AD is equal to EM, MD ; but EM, MD are greater
(20. 1.) than ED ; therefore also AD is greater than ED : again, be-
cause ME is equal to MF, and MD common to the triangles EMD,
FMD: EM, MD are equal to FM, MD ; D

but the angle EMD is greater than the
angle FMD ; therefore the base ED is
greater (24. 1.) than the base FD: in
fike maimer it may be shown that FD is
greater than CD : therefore DA is
the greatest: and DE greater than
DF, and DF than DC: and because
MK^ KD are greater (20. 1.) than
MD, and MK is equal to MG, the
remainder KD is greater (4 Ax.) than
the remainder GD, that is, GD is
less than KD: and because MK, DK, C L-
are drawn to the point K within the
triangle MLD, from M, D, the ex-
tremities of its side MD; MK, KD
are less, (2L 1.) than ML, LD whereof
MK is equal to ML; therefore the re-
mainder DK is less than the remainder
DL : in like manner it may be shown that DL is less than DH ;
therefore DG is the least, and DK less than DL, and DL than DH ;
also there can be drawn only two equal straight lines from the point
D to the circumference, one upon each side of the least; at
the point M, in the straight line MD, make the angle DMB equal

BOOK III,

THB «L£ME1IT9 OF VQCUD,

to the angle DMK, and join DB; and because MK is equa) to MB»
and MD common to the triangles KMD, BMD, tbe two sides KM,
MD are equal to the two BM, MD ; and the angle EMD is equaj to
the angle BMD ; therefore the base DK is equal to (4. 1.) the oase
DB : but) besides DB, there can be no straight line drawn from D
to the circumference equal to DK : for, if there can, let it be DN ;
and because DK is equal to DN, and also to DB ; therefore DB is
equal to DN^ that is, the nearer to the least equal to the more remote,
which is impossible. If, therefore, any point, &c. Q. K D.

PROP. IX. THEOR.

If a point be taken within a circle, from which there fall more
than two equal straight lines to the circumference, that point is
the centre of the circle.

Let the point D be taken within the circle ABC, from which to
the circumdference there fall more than two equal straight Imes, viz.
DA, DB, DC : the point D is the centrev of the circle.

For, if not, let E be the centre, join
DE, and produce it to the circumference
in F, G ; then FG is a diameter of the
circle ABC : and because in FG, the di-
ameter of the circle ABC, there is taken
the point D which is not the centre, DG
shall be the greatest line from it to the
circumference, and DC greater (7. 3.)
than DB, and DB than DA ; but they
are likewise equal, which is impossible : A B

therefore E is not the centre of the circle ABC : in like manner it
may be demonstrated, that no other point but D is the centre ; D
therefore is the centre. Wherefore, if a point be taken, &c. (J. E. D.

PROP. X. THEOR.

One circumference of a circle cannot cut another in more
than two points.

If it be possible, let the circumfe- A

rence FAB cut the circumference DEF
in more than two points, viz. in B, G,
F ; take the centre K of the circle ABC,
and join KB, KG, KF ; and because
within the circle DEF there is taken
the point K, from which to the circum-
ference DEF fall more than two equal
straight lines KB, KG, KF, the pomt
K is (9. 3.) the centre of the circle
DEF : but K is also the centre of the
circle ABC ; therefore the same point ^

is the centre of two circles that cut one another which is impossible

THB BLEMEirrS OF EUCLID.

BOOK If I.

(5. 3.) Therefore one circumference of a circle cannot cut another
in more than two points. Q,. £l. D.

PROP. XL THEOR.

If two circles touch each other internally' the straight line
which joins their centres being produced shall pass through the
point of contact.

Let the two circles ABC, ADE touch each other internally in the
point A, and let F be the centre of the circle ABC, and G the centre
of the circle ADE : the straight line which A

joins the centres F, G, being produced,
passes through the point A.

For, if not, let it fall otherwise, if possi- H
ble, as FGDH, and join AF, AG : and be-
cause AG, GF are greater (20. 1.) than FA,
that is than FH, for FA is equal to FH,
both being from the same centre; take
away the common part FG ; therefore the
remainder AG is greater than the remain-
der GH : but AG is equal to GD ; there-
fore GD is greater than GH, the less than the greater, which is im-
possible. Therefore the straight line which joins the points F, G
cannot ML otherwise than upon the point A, that is, it must pass
through it. Therefore, if two circles, &c. d. E. D.

PROP. Xn. THEOR.

If two circles touch each other externally» the straight line
which joins their centres shall pass through the point of contact

Let the two circles ABC, ADE touch each other externally in the
point A ; and let F be the centre of the circle ABC, and G the centre
of ADE: the straight Me which joins the points F, G shall pass
through the point of contact A.

For, if not, let it pass otherwise, if possible, as FCDG, and join
FA, AG : and because F is the centre of the circle ABC, AF is equal
to FC : also because G is the ^— -^ E

centre of the circle ADE, AG
is equal to GD: therefore
FA, AG are equal to FC,
DG ; wherefore the whole FG
is greater than FA, AG : but
it is also less (20. L); which
is impossible: therefore the
straight line which joins the
points F, G shall not pass otherwise than through the point of con-
tact A, that is, it must pass through it. Therefore, if two circles,
^. €L E. D.

BOOK lit.

TRB BLBMBNTS OP BOCLID.

PROP. Xin. THEOR.

One circle cannot touch another in more points than one,
whether it touches it on the inside or outside.*

For, if it be possible, let the circle EBF touch the circle ABC m
more points than one, and first on the inside, in the points B, D ;
join ED, and draw (10. 11. 1.) GH bisecting ED at right angles.
Therefore, because the pomts E, D are in the circumference of

each of the circles, the straight line ED fells within each (2. 3.) of
them: and their centres are (Cor. 1. 3.) in the straight lineGH
which bisects BD at right angles; therefore GH passes through
the point of contact (11. 3.); but it does not pass through it, be-
cause the points E,.D are without the straight lineGH, which is
absurd : therefore one circle cannot touch another on the inside in
jnore points than one.

Nor can two circles touch one another on the outside in more
than one x)oint : for, if it be possible, let the circle ACE touch the
circle ABC in the points A, C, and join AC: therefore, because
the two points A, C are in the circumference
of the circle ACK, the straight line AC
which joins them shall fall within (2. 3.)
the circle ACE: and the circle ACE is
without the circle ABC ; and therefore the
straight line AC is without this last circle ;
but because the points A, C are in the cir-
cumfqrence' of the circle ABC, the straight
line AC must be within (2. 3.) the same cir-
cle, which is absurd; therefore one circle
cannot touch another on the outside in more
than one point : and it has been shown that
they cannot touch on the inside in more
points than one. Therefore one circle, &c.
Q. E.D. ♦

PROP. XIV. THEOR. *

Equal straight lines in a circle are equally distant from the
centre ; and those which are equally distant ifrom the centre are
equal to one another.

* See Note.

Qfil TSB BLBMBNTS QP BUCUD. BOOK lU.

Let the straight lines AB, CD, in the drcle ABDC, be equal to one
another : they are equally distant from the centre.

Take E the centre of the circle ABDC, and from it draw EF*
EG perpendiculars to AB, CD; then, because the straight line

EF, passing through the centre, cuts the straight line AB, which
does not pass through the centre, at right
angles, it also bisects (3. 3.) it : wherefore
AF is equal to FB, and AB double of AF«
For the same reason, CD is double of CG ;
and AB is equal to CD ; therefore AF is
equal to CG : and because AE is equal to
EC, the square of AE is equal to the
square of EC ; but the squares of AF, FE
are equal (47. 1.) to the square of AE, be-
cause the angle AFE is a right angle;
and, for the like reason, the squares of

EG, GC are equal to the square of EC : therefore the squares of
AF, FE are equal to the squares of CG, GE, of which the squ^e
of AF is equal to the square of CG, because AF is equal to CG ;
therefore the remaining square of FE is equal to the remaining^
square of EG» and the straight line EF is th^efore equal to EG :
but straight lines in a circle are said to be ^equally distant from the
centre, when the perpendiculars drawn to them from the centre
are eqjuai (4. def. 3.) therefore ABCD are equally distant from the
oentre.

Next, if the straight lines AB, CD be equally distant from the
centre ; that is, if FE be equal to EG, AB is equal to CD : for, the
some construction being made, it may, as before, be demonstrated
that AB is double of AF, and CD double of CG, and that the
squares of EF, FA are equal to the squares of EG, GC ; of which
the square of FE is equal to the square of EG, because FE is
equal to EG ; therefore the remaining square of AF is ^ual to the
remaining square of CG; and the straight line AF is therefore
equal to CG: and AB is double of AF, and CD double of CG;
wherefore AB is equal to CD. Therefore equal straight lines, &c«

a. E.D.

PROP. XV. THEOR.

The diameter is the greatest straight line in a circle ; and, of
all others, that which is nearer to the centre is always greater
than one more remote ; and the greater is nearer to the centre
than the less.*

« * See Note.

BOOK in. TRB CLBMBNTS 01* BOCLID.

Let ABCD be a circle, of which the dia- A B

meter is AD, and the centre E; and let BO
be nearer to the centre than FG ; AD is greater
than any straight line BO which is not a dia-
meter, and BC greater than FG.

From the centre draw EH, EK perpendicu-
lar to BC, FG, and join EB, EC, EF ; and
because AE is equal to EB, and ED to EC,
AD is equal to EB, EC ; but EB, EC are
greater (20. 1.) than BC ; wherefore also AD
is greater than BC.

And, because BC is nearer to the centre than FG, EH is less (5.
def. 3.) than EK; but, as it was demonstrated in the preceding, BO
is double of BH, and FG double of FK; and the squares of EH, HB
are equal to the squares of EK, KF, of which the square of EH is
less than the square of EK, because EH is less than EK ; therefore
the square of BH is groater than the square of FK, and the straight
llne'fiH greater than FK; and therefore BC is greater than FG.

.Next, let BC be greater than FG ; BC.is nearer to the centre than
FG, that is, the same construction being made, EH is less than EK:
because BC is greater than FG, BH likewise is greater than KF : and
the squares of BH, HE are equal to the squares of FK, KEI, of which
the square of BH is greater than the square of FK, because BH is
greater than FK ; therefore the square of EB is less than the square
of EK, and the straight line EH less than EK. Wherefore the dia»
meter, &c. Q. K D.

PROP. XVI. THEOR.

Thb straight line drawn at right angles to the diameter of a
circle, from the extremity of it, fells without the circle ; and no
straight line can be drawn between that straight line and the
circumference from the extremity, so as not to cut the circle ; or
which is the same thing, no straight line can make so great an
acute angle with the diameter at its extremity, or so small an
angle with the straight line which is at right angles to it as not^
to cut the circle.*

Let ABC be a circle, the centre of which is D, and the diameter
AB; the straight line drawn at right, angles to AB from its extremity
A, shall M without the circle.

* See Note.

THE ELEMSNTS OF EUCLID.

BOOK UI.

For, if it does not, let it fall, if possi-
ble, within the circle, as AC, and draw
DC to the point C where it meets the cir-
cumference ; and because DA is equal to
DC, the angle DAC is equal (5. 1.) to the g
angle ACD; but DAC is a right angle,
therefore ACD is a rigiit angle, and the
angles DAC, ACD are therefore equal to
two right angles; which is impossible
(17. !.)• therefore the straight line drawn
from A at right angles to BA does not fall within the circle; in the
same manner, it may be demonstrated, that it does not fall upon the
circumference ; therefore it must fall without the circle, as AE.

And between the straight line AEand the circumference no straight
line can be drawn from the point A which does not cut the circle : for,
if possible, let FA be between them, and from the point D draw (12.
1.) DO perpendicular to FA, and let it meet Ihe circumference in H:
and because AGD is bright angle, and EAG less (19. 1.) than a right
angle: DA is greater (19. 1.) thanDG: but DA is equal to DH :
therefore DH is greater than DG, the less E

than the greater, which is impossible:
therefore no strsdght line can be drawn
from the point A between AE and the cir-
cumference, which does not cut the circle;
or, which amounts to the same, thing,
however great an acute angle a straight
line makes with the diameter at the point
A, or however small an angle it makes B |
with AE, the circumference passes be-
tween that straight line and the perpen-
dicular AE. 'And this is all that is to be
understood, when, in the Greek text, and
translations from it, the angle of the semi-
circle is said to be greater than any acute rectilineal angle, and the
remaining angle less than any rectilineal angle.'

Cor. From this it is manifest, that the straight line which is
drawn at right angles to the diameter of a circle from the extremity
«of it, touches the circle ; and that it touches it only in one point, be-
cause, if it did meet the circle in two, it would fall within it (2. 3.)
' Also it is evident that there can be but one straight line which
touches the circle in the same point.'

PROP. XVn. VROB.

To draw a straight line from a given point, either without or
in the circumference, which shall touch a given circle.

First, let A be a given point without the given circle BCD : it is
required to draw a straight line from A which shall touch the cirde.

BOOK ni.

THE SLEHfiNTS OF BDCLID.

Find (1. 3.) the centre £ of the circle, and join AE; and from the
centre £, at the distance £A, describe the curcle AFG ; from the point
D draw (11. 1.) DF at right angles to EA, and join EBF, AB. AB
touches the circle BCD.

Because £ is the centre of the
circles BCD, AFG, £IA is equal to
£F; and £D to £B; therefore the
two sides A£, £B are equal to the
two FE, ED, and they contain the
angle at E common to the two tri- G
angles AEB, FED; therefore the
base DF is equal to the base AB,
and the triangle FED to the trian-
gle AEB, and the other angles to
the other angles (4. 1.); therefore
the angle EBA, is equal to the angle EDF: but EDF is a right angle,
wherefore EBA is a right angle ; and EB is drawn from the centre ;
but a straight line drawn from the extremity of a diameter, at right
angles to it, touches the circle (Cor. 16. 3.) : therefore AB touches
the circle ; and it is drawn from the given pomt A. Which was to
be done.

But, if the given point be in the circumference of the circle, as the
point D, draw DE to the centre E, and EF at right angles to DE;
DF touches the circle (Cor. 16. 3.).

PROP. XVIIL THEOR.

If a straight line touch a circle^ the straight line drawn from
the centre to the point of contact, shall be perpendicular to the
line touching the circle.

Let the straight line DE touch the circle ABC in the point C ; take
the centre F, and draw the straight line FC ; FC is perpendicular to
DE.

For, if it be not, from the point F draw FBG perpendicular to DE;
and because FGC is a right angle, GCF A

is (17. 1.) an acute angle; and to the
greater angle the greatest (19. 1.) side is
opposite; therefore FC is greater than
PG, but FC is equal to FB; therefore FB
is greater than FG, the less than the
greater, which is impossiUe; wherefore
FG is not perpendicular to DE : in the
same manner it may be shown, that no
other is perpendicular to it besides FC,
that is, FC is perpendicular to DE. There- jy
fore, if a straight line, &c. Q. E. D.

THE ELEMENTS OF EUdLID.

Boo)c in.

iPROP. XIX. THEOft.

If a straight line touch a circle, and from the point of contact
a straight line be drawn at right angles to the touching line, the
centre of the circle shall be in that line.

Let the straight Une DE touch the circle ABC in C, and fr6m C let
CA be drawn at right angles to t)K; the centre of the circle is in CA.

For, if not, let F be the centre if possible ; and join CF : because
DE touches the circle ABC, and FC A

is drawn from the centre to the point
of contact, FC is perpenpicular (18.
3.) to DE ; therefore FCE is a right
angle ; but ACE is also a right angle ;
therefore the angle FCE is equal to
the angle AClS, the Jess to the great-
er, which is impossible; wherefore F
is not th^ centre of the circle ABC ;
, in the same manner it may be shown,

that no other point which is not in ^ ^ ^

CA, is the centre; that is, the centre
is in CA. Therefore if a straight line, &c. Q. E. D.

PROP. XX. THEOR.

The angle at the centre of a circle is double of the angle at
the circumference, upon the same base, that is, upon the same
'|)'art of the circumference.*

Let ABC be a circle, and BEC an angle at the centre, and BAG
yn angle kt the circumference, which have the same circumference
' BC for their base ; the angle 6EC is double of A

th^ angle BAC.

First, let E the centre of the circle be with-
in the angle BAC, and join AE, and produce
it to F ; because EA is equal to EB, the angle / / I \ \

EAB is equal (6. L) to the angle EBA ; there-
fore the angles EAB, EBA are double of the
angle EAB, but the angle BEF is equal (32.
L) to the angles EAB, EBA; therefore also «
the angle BEF is double of the angle EAB :
for the same reason, the angle 1?'EC is double F

of the angle EAC : therefore the whole angle^ BEC is double of the
whde ailgle BAC.

« See Note.

BOOK lU.

THS BLSMSNTS OP SUCLID^

Cff

Again ; ]et £ the. centre of the circle
be without the angle BDC, and join DE
and produce it to G. It may be demon-
strated, as in the first case, that the angle
GEO is double of the angle GDC, and
that GEB a part of the first is double of
GDB a part of the other ; therefore the
remaining angle BEC is double of the p
remaming angle BDC. Therefore the
angle at the centre, &c. a. E. D.

PROP. XXI. THEOR.

Thc angles in the same segment of a circle are equal to one
another.*

Let ABCD be a circle, and BAD, BED
angles in the same segment BAED : the
an^es BAD, BED are equal to one another.

Take F the centre of the circle ABCD ;
and, first, let the segment BAED be great-
er than a semicircle, and join BF, FD : and
because the angle BFD is at the centre,
and the angle BAD at the circumference,
and that they have the same part of the
circtimference, viz. BCD, for their base ;
therefore the angle BFD is double (20. 3.)
of the angle BAD : for the same reason, the angle BFD is double of
the angle BED : therefore the angle BAD is equal to the angle BED.

But, if the segment BAED be not greater than a semicircle, let

BAD, BED be angles in it; these also
are equal to one another : draw AF to
the centre, and produce it to C, and join
CE: therefore the segment BADC is
greater than a semicircle; and the angles
in it, BAC, BEC are equal, by the first
case ; for the same reason because CBED
is greater than a semicircle, the angles
CAD, CED are equal : therefore the whole
angle BAD is equal to the whole angle
BED. Wherefore the angles in the same
segment, d:.c. d. *K D.

PROP. XXII. THEOR.

Thb opposite angles of any quadrilateral figure describe^ in
a circle, are together equal to two right angles.

» See Note.

eS THE ELEMSirro OP EUCLID. BOOK HI,

Let ABCD be a quadrOateral figure in the circle ABCD ; any two
of its opposite an^es are together equal to two right angles.

Join AC, BD ; and because the three angles of every triangle are
equal (32. 1.) to two right angles, the three angles of the triangle
CAB, viz. tiie angles CAB, ABC, BCA are equal to two right angles :
but the angle CAB is equal (21. 3.) to the JD

angle CDB, because they are in the same
segment BADC, and the angle ACB is
equal to the angle ADB, because they are
in the same segment ADCB : therefore the
whole angle ADC is equal to the angles
CAB, ACB: to each of these equals add A
the angle ABC : therefore the angles ABC,
CAB, BCA are equal to the angles ABC,
ADC : but ABC, CAB, BCA are equal to
two right angles ; therefore also the angles ABC, ADC are equal to
two right angles ; in the same manner, the angles BAD, DCB may
be shown to be equal to two right angles. Therefore the opposite
angles, &c. d. E. D.

IPROP. XXIII. THEOR.

^ Upon the same straight line, and upon the same side of it,
there cannot be two similar segments of circles, not coinciding
with one another.*

If it be possible, let the two similar segments of circles, viz. ACB,
ABD be upon the same side of the same straight line AB, not com-
ciding with one another: then, because the circle ACB cuts the

circle ADB in the two points A, B, they cannot ^ D

cut one another in any other point (10. 3.) : one
of the segments must therefore fall within the
other ; let ACB fall within ADB, and draw the
straight line BCD, and join CA, DA : and be-
cause the segment ACB is similar to the seg- A B
ment ADB, and that similar segments of circles contain (11. def. 3.)
equal angles : the angle ACB is equal to the angle ADB, the exterior
to the interior, which is impossible (1 6. 1.). Therefore, there cannot
be two similar segments of a circle upon the same side of the same
line, which do not coincide. €t. E. D.

PROP. XXIV. THEOR.

Similar segments of circles upon equal straight lines, are equal
to one another.*

Let AEB, CFD be similar segments 'of circles upon the equal
straight lines AB, CD : the segment AEB is equal to the segment
CFD.

* See Notes.

B0OK« III.

THE SLBMBHTS OF KUCLW.

For, if the segment E F

A£B be applied to the
segment CFD, so as the
point A be on G, and the
straight line AB upon A B C D

CD, the point B shall coincide with the point D, because AB is equal
to CD : therefore the straight line AB coinciding with CD, the seg-
ment AEB must (23. 3.) coincide with the segment CFD, and there-
fore is equal to it. Wherefore similar segments, &c. Q. E. D.

PROP. XXV. PROB.

A SEGMENT of a circlc being given to describe the circle of
which it is the segment*

Let ABC be the given segment of a circle ; it is required to de-
scribe the circle of which it is the segment.

Bisect (10. 1.) AC in D, and from the point D draw (11. 1.) DB
at right angles to AC, and join AB; first, let the angles ABD, BAD,
be equal to one another ; then the straight line BD is equal (6.*1.) to

DA, and therefore to DC, and because the three straight lines DA,

DB, DC, are all equal ; D is the centre of the circle (9. 3.) : from the
centre D, at the distance of any of the three DA, DB, DC, describe
a circle ; this shall pass through the other points ; and the circle of
which ABC is a segment is described : and because the centre D is
in AC ; the segment ABC is a semicircle : but if the, angles ABD,

B B

BAD are not equal to one another, at the point A, in the straight
line AB, make (23. 1.) the angle BAE equal to the angle ABD, and
produce BD, if necessary, to E, and join EC : and because the angle
ABE is equal to the angle BAE, the straight line BE is equal (6. 1.)
to EA ; and because AD is equal to DC, and DE common to the tri-
angles ADE, CDE, the two sides AD, DE are equal to the two CD,
DE, each to each ; and the angle ADE is equal to the angle CDE,
for each of them is a right angle ; therefore the base AE is equal
(4. 1.) to the base EC : but AE was shown to be equal to EB, where-
fore also BE is equal to EC : and the three straight lines AE, EB,
EC are therefore equal to one another ; wherefore (9. 3.) E is the
centre of the circle. From the centre E, at the distance of any of
the three AE, EB, EC, describe a circle, thiis shall pass through the

* See Note.

70 TH9 SLBMBurpa or euclid. book Ul.

other points ; and the circle of which ABC is a segment is described :
and it is evident, that if the angle ABD be greater than the angle
BAD, the centre E falls without the segment ABO, which therefore
is less than a semicircle; but if the angle ABD be less than BAD,
the centre E falls within the segment ABC, which is therefore great-
er than a semicircle : wherefore a segment of a circle being given,
the circle is described of which it is a segment. Which was to be
done.

PROP. XXVI. THEOR.

In equal circles, equal angles stand upon equal circumferences,
whether they be at the centres or circumferences.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF
at their centres, and BAC, EDF at their circumferences : the circum-
ference BKC is equal to the circumference ELF.

Join BC, EF ; and because the circles ABC, DEF are equal, the
straight lines drawn from their centres are equal : therefore the two
sides BG, GO are equal to the two EH, HF ; and the angle at G is

C E

eqnal to the angle at H ; therefoie the base BC is equal (4. L) to the
base EF ; and because the angle at A is equal to the angle at D, the
segment BAC is similar (IL def 3.) to the segment EDF; and they
are upon equal straight lines BC, EF ; but similar segments of cir-
cles upon equal straight lines are equal (24. 3.) to one another ; there-
fore the segment BAC is equal to the segment EDF ; but the whole
circle ABC is equal to the whole EDF ; therefore the remaining seg-
ment BKC is equal to the remaining segment ELF, and the circum-
ference BKC to the circumference ELF. Wherefore, in equal cir-
cles, &c. d. E. D.

PROP. XXVII. THEOR.

In equal circles, the angles which stand upon equal circum-
ferences are equal to one another, \yhether they be at the centres
or circumferences. '

Let the angles BGC, EHF at the centres, and BAC, EDF, at
the ch*cumference of the equal cirdes ABC, DEF, stand upon the

BOOK 111.

THE BLEMENTS OP EUCLID.

ec^ual cttcumferences BC, £F ; the angle BGC Is equal to the angle
EHF, and the angle BAG to the angle EDF.

If the angle BQO be equal to the angle EHF, it is manifest (20. 3.)
that the angle BAG is also equal to EDF: but, if not, one of tlrem is

the greats ; let BGrC te the greater; and at the pijint G, in the
straight line BG, make (23. I.) the angle BGK equal to the angle
EHF ; but equal angles stand upon equal circumferences (26. 3.)
when they are at the centre; therefore the circumference BK is
equal to the circumference EF : but EF is equal to BC ; therefore
also BK is equal to BG, the less to the greater, which is impossible :
therefore the angle BGC is not unequal to the angle EHF ; that is, it
is equal to it : and the angle at A is half of the angle BGC, and the
angle at D half of the angle EHF : therefore the angle at A is equal
to the angle at D. Wherefore, in equal circles, &c. €t E. D.

PROP. XXVra. THEOR.

Iif equal circles, equal straight lines cut off equal circumfe-
rences, the greater equal to the greater, and the less to the less.

Let ABC, DEF be equal circles, and BC, EF equal straight lines
in them, which cut off the two greater circumferences BAG, EDF,
and the two less BGC, EHF ; the greater BAG is equal to the great-
er EDF, and the less BGC to the less EHF.

Take (1. 3.) K, L, the centres of the circles, and join BK, KG, EL,
LF : and because the circles are equal, the straight lines from their

A D

centres are equal : therefore BK, KG are equiea to EL, LF; and the
btte BC is equal to the base EF; therefore the angle BKG is equal
^.hy to theangic ®LF : but equal angles stattd iqKKi eqOal (26, 3.)

72 THE ELEMEirrS OF EUCLID. EOOK UI.

clrcuinferences, when they are at the centres^; there^re the circum-
ference BGC is equal to the circumference EHF. But the whole circle
ABC is equal to the whole EDF ; the remaining part therefore of the
circumference, viz. BAG, is equal to the remaining part EDF. There-
fore, in equal circles, ^. Q,. E. D.

PROP. XXIX. THEOR.

Ii7 equal circles equal circumferences are subtended by equal
straight lines.

Let ABC, DEP be equal circles, and let the circumferences BGC,
EHF also be equal ; and join BC, EF : the straight liner BC is equal
to the straight line EF.

Take (1. 3.) K, L, the centres of the circles, and join BK, KC, EL,
LF : and because the circumference BGC is equal to the circumfe-

A D

C E

G H

rence EHF, the angle BEC is equal (27. 3) to the angle ELF : and
because the circles ABC, DEF are equal, the straight lines from their
centres are equal : therefore BK, KC are equal to EL, LF, and they
contain equal angles : therefore the base BC is equal (4. 1.) to the
base EF. Therefore, in equal circles, &c. d. E, D.

PROP. XXX. PROB.

To bisect a given circumference^ that is, to divide it into two
equal parts.

Let ADB be the given circumference, it is required to bisect it.

Join AB, and bisect (IQ. L) it in C ; from the point C draw CD at
right angles to AB, and join AD, DB: the circumference ADB is bi-
sected in the point D.

Because AC is equal to CB, and CB common to the triangles
ACD, BCD, the two sides AC, CD are equal D

to the two BC, CD ; and the angle ACD is
equal to the angle BCD, because each of
them is a right angle ; therefore the base AD
is equal (4. 1.) to the base BD: but equal A C B

straight lines cut off equal (28. 3.) circumferences, the greater equal
to the greater, and the less to the less, and AD, DB are each of
them less than a semicircle; because DC passes through the cen-
tre (Cor. L 3.); wherefore the circumference AD is equ^ t9 the cir-

BOOK III. THB BLniBNTS OF BUCLID. 7S

cumference DB : therefore the given circumferenoe is bisected in D.
Which was to be done.

PROP. XXXI. THEOR.

Ill a circle, the angle in a semicircle is a right angle ; but the
angle in a segment greater than a semicircle is less than a right
angle; and the angle in a segment less than a semicircle ib
greater than a right angle.

Let ABCD be a circle, of which the diameter is BC, and centre
E ; and draw CA, dividing the circle into the segments ABC, ADC,
and join BA, AP, DC ; the angle in the semicircle BAC is a right
angle ; and the angle in the segment ABC, which is greater than a
semicircle, is less than a right angle ; and the angle in the segment
ADC, which is less than a semicircle, is greater than a right angle.

Join AE, and produce BA to F; and because BE is equal to EA,
the angle EAB is equal (5. 1.) to EBA ; also, because A£ is equal
to EC, the angle EAC is equal to ECA ; F

wherefore the whole angle BAC is equal to
the two angles ABC, ACB ; but FAC, the
exterior angle of the triangle ABC, is equal

S3. 1.) to the two angles ABC, ACB;
erefore the angle BAC is equal to the
angle FAC, and each of them is therefore
a right (10. deC 1.) angle ; wherefore the
angle BAC in a semicircle is a right angle.

And because the two angles ABC, BAC,
of the triangle ABC are together less (17.
1.) than two right angles, and that BAC
is a right angle, ABC must be less than a right angle: and there^
fore the angle in a segment ABC greater than a semicircle, is lesa
than a right angle.

And because ABCD is a quadrilateral figure in a circle, any two
of its opposite angles are equal (22. 3.) to two right angles ; there-
fore the angles ABC, ADC are equal to two right angles ; and ABC.
is less than a right angle ; wherefore the other ADC i» greater than
a right angle.

Besides, it is manifest, that the circumference of the greater seg-
ment ABC fells without the right angle CAB, but the circumference
of the less segment ADC Ms within the right angle CAF. «< And
this is all that is meant, when in the Greek text, and the translationa
from it, the angle of the greater segment is said to be greater, and
the ang^e of the less segment is said to be less, than a right angle."

Cor. Frpm this it is manifest, that if one angle of a triangle be
equal to the other two, it is a r^t angle, because the angle adjacent
to it is equal to the same two } and when the adjacent angles are
equal, they afe right angles.

74- THB BLEMENTS OF EUCLID. BOOK. III.

PROP. XXXn. THEOR.

Ip a straight line touch a circle, and from the point of contact
a straight line be drawn cutting the circle, the angles made by
this line with the line touching the circle, shall be equal to the
angles which are in the alternate segments of the circle.

Let the straight line EP touch the circle ABCD in B, and from the
point B let the straight line BD be drawn, cutting the circle : The
angles which DB makes with the touching line EF ^all be equal to
the angles in;the alternate segments of the circle : that is, the angle
FED is equal to the angle which is in the segment DAB, and the an-
gle DBE to the angle in the segment BCD.

From the point B draw (11. 1.) BA at right angles to EP, and
take any point C in the circumference BD ; and join AD, DC, CB ;
and because the straight line EF touclies the circle ABCD in the
point B, and BA is drawn at right A

angles to the touching line from
the point of contact B, the centre
of the circle is (19. 3.) in BA;
therefore the angle ADB in a semi-
circle is a right (31. 3.) angle, and
consequently the other two angles
BAD, ABD are equal (32. 1.) to a
right angle : but ABF is likewise a
right angle; therefore the angle
ABF is equal to the angles BAD, E B F

ABD : take from these equals the common angle ABD ; therefore
the remaining angle DBF is equal to the angle BAD, which is in the
alternate segment of the circle : and because ABCD is a quadri-
lateral figure in a circle, the opposite angles BAD, BCD are equal
(22. 3.) to two right angles ; therefore the angles DBF, DBE, being
likewise equal (13. 1.) to two right angles, are equal to the angles
BAD, BCD ; and DBF has been proved equal to BAD : therefore the
remaining angle DBE is equal to the angle BCD in the alternate
segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXXin. PROB.

Upon a given straight line to describe a segment of a circle,
containing an angle equal to a given rectilineal angle.*=

Let AB be the given straight line, and the angle at C the given
rectilineal angle ; it is required to describe upon the given straight
Kne AB a segment of a circle, containing an angle equal to the
angle C.

* See Note.

BOOK III.

THE SLIMBKT8 OF BUCLID.

First, let the angle at G be
a right angle, and bisect (10.
1.) A6 in F, and from the cen-
tre F, at the distance FB, de-
scribe the semicircle AHB;
therefore the angle AHB in a
semicircle is (31. 3.) equal to
the right angle at C.

But, if the angle C be not a right angle, at the point A, in the
straight line AB, make (23. 1.) the angle BAD equal to the angle C,
and fcom the point A draw (1 1.1.) H

AE at right angles to AD : bisect
(10. 1.) AB in F, and from F draw
(11. 1.) FG at right angles to AB,
and join GB: and because AF is
equal to FB, and FG common to
the triangles AFG, BFG, the two
sides AF, FG are equal to the two
BF, FG; and the angle AFG is
equal to the angle BFG ; therefore
the base AG is equal (4. 1.) to the
base GB ; and the circle described
from the centre G, at the distance GA, shall pass through the^point
B; let this be the circle AHB : and because from the point A the ex-
trenuty of the diameter AE, AD is drawn at right angles to AE,
therefore AD (Cor. 16. 3.) touches the circle ; and because AB drawn
from the point of contact A cuts C
the circle, the angle DAB is equal
to the angle in the alternate segment
AHB (32. 3.) : but the angle DAB
is equal to the angle C, therefore
also the angle C is equal to the an-,
gle in the segment AHB : where-
fore upon the given straight line AB
the segment AHB of a circle is de-
scribed ,which contains an angle equal to the given angle at G.
Which was to be done.

PROP. XXXIV. PROB.

To cut off a segment from a given circle which shall contain
an angle equal to a given rectilineal angle.

Let ABC be the given circle, and D the given rectilineal angle ; it
is required to cut off a segment from the circle ABC that shsdl con-
tain an angle equal to the given angle D.

Draw (17. 3) the straight line EF touching the circle ABC in the

THB ELBBUSMT* OP BUCUD.

BOOJL UU

point B, and at the point B, in the
straight line BF, make (23. 1.) the
angle FBC equal to the angle D ;
therefore, because the straight line
EF touches the circle ABC, and
BC is drawn from the point of con-
tact B, the angle FBC is equal (32. D
3.) to the angle in the alternate seg-
ment BAC of the circle : but the

angle FBC is equal to the angle D; e B F

therefore the angle in the segment BAC is equal to the angle D :
wherefore the segment BAC is cut off from the given circle ABC
containing an angle equal to the given angle D. Which was to be
done.

PROP. XXXV. THEOR.

If two straight lines within a circle cut one another, the rect-
angle contained by the segments of one of them, is equal to the
rectangle contained by the segments of the other.^

Let the two straight lines AC BD, within the cu-cle ABCD,
cut one another in the point E : the rectangle contained by AE,
EC is equal to the rectangle contained by BE,
ED.

If AC, BD pass each of them through the
centre, so that E is the centre ; it is evident,
that AE, EC, BE, ED, bemg all equal, the
rectangle AE, EC is likewise equal to the rect-
angle BE, ED.

But let one of them BD pass through the centre, and cut the
other AC, which does not pass through the centre, at right angles,
in the point E ; then, if BD be bisected in F, F is the centre of the
circle ABCD; join AF: and because BD which passes through
the centre, cuts the straight line AC, which does not pass through
the centre, at right angles in E, AE, D

EC are equal (3. 3.) to one another:
and because the str£dght line BD is cut
into two equal parts in the point F, and
into two unequal in the point E, the
rectangle BE ED, together with the
square of BF, is equal (5. 2.) to the
square FB; tiiat is, to the square of A
FA; but the squares of AE, EF are
equal (47. 1.) tb the square of FA;
therefore the rectangle BE, ED, to- B

gether with the square of EF, is equal to the squares of AE, EF:
take away the common square of EF, and the remaining rectangle

* See Note.

BdoK ni.

TRi lEAMnrrt or bvcud.

BB» BX> te equal to the renudnliig square erf" AE; tiiot la, to the rect-
angle AE, EC.

Next, let BD, which passes through the centre, cut the other AC,
which does not pass through the centre, In E, but not at right angles :
then, as before, if BD be bisected in F, F is the centre of the circle.
Join AF, and from F draw (12. 1.) FQ perpendidilar to AC ; there-
fore AG is equal (3. 3.) to GD ; wherefore the rectangle AE, EC, to*
gether with the square of EG, is equal (5. 2.) to the square of AG : to
each of these equals add the square of GF ; therefore the rectangle
AE, EC, together with the squares of EG, GF, is equal to the squares
of AG, GF : but the squares of EG, GF are jx
equal (47. 1.) to the square of EF, and the
squares of AG, GF are equal to the square
of AF ; therefore the rectangle AE, EC, to-
gether with the square of EF, is equal to
the square of AF ; that is; to the square of
FB: but the square of FB is equal (6. 2.) to A
the rectangle BE, ED, together with the
square oi EF : therefore the rectangle AE,
EC, togeth^ with the square of EF, is equal to the rectangle BE, DE,
together with the square of EF: take away the comm<» square of
EF, and the remaining rectangle AE, EC is therefore equal to the
remaining rectangle BE, ED.

Lastly, let neither of the straight lines AC, BD pass through th«
centre: take the centre F, and through £, < i

the intersection of the straight lines AC,
DB draw the diameter GEFH : and because

the rectangle AE, EC is equal, as has been / X t / D

shown, to the rectangle GE, EH: and, for I x r / ^

the same reason, the rectangle BE, ED is *

equal to the same rectangle GE, EH; there- ^^^ /v / C

fore the rectangle AE, EC is equal to the
rectangle BE, ED. Wherefore, if two
straight lines, &c. Q. E. D.

PROP. XXXVL THEOR.

If from any point without a circle two strai^t lines be drawn,
one of which cuts the circle, and the other touches it ; the rect-
angle containi^ by th6 whole line which cuts the circle, and the
part of it without tte circle, shall be equal to the square of the

line which touches it.

•

Let D be any pohit without the circle ABC, and DGA, DB two
straight lines drawn from it, of which DCA cuts the circle, and DB
touches the same ; the rectangle AD, DC is equal to the square of DB.

7»

THE ELEHEHTS OS EOCE.ID.

BOOK III*

Bather DCA pksses through the centre, or it does not j first, let it
pass through the centre E, and join EB ; therefore the angle EK> ia
a right (18. 3.) angle : and because the straight D

line AC is bisected in E, and produced to the
point D, the rectangle AD, DC, together with
the square of EC, is equal (6. 2.) to the square
of ED, and CE is equal to EB: therefore the
rectangle AD, DC, together with the square of
EB, is equal to the square of ED: but the
square of ED is equal (47. I.) to the squares
of EB, BD because EBD is a right angle:
therefore the rectangle AD, DC, together with
the square of EB, is equal to the squares of
EB, BD : take away the common square of
EB; therefore the remaining rectangle AD,
DC is equal to the square of the tangent DB.

But if DCA does not pass through the centre of the circle ABC,
take (1. 3.) the centre E, and draw EP perpendicular (12. 1.) to AC,
and join EB, EC, ED : and because the straight line EF, which passes
through the centre, cuts the straight line AC, which does not pass
through the centre, at right angles, it shall d

likewise bisect (3. 3.) it; therefore AF is equal
to FC : and because the straight line AC is
bisected in F, and produced to D, the rectan-
gle AD, DC, together with the square of FC,
is equal (6. 2.) to the square of FD : to each
of these equals add the square of FE ; there-
fore the rectangle AD, DC, together with the
squares of CF, FE, is equal to the squares of
DF, FE : but the square of ED is equal (47. 1.)
to the squares of DF, FE, because EFD is a
right angle: and the square of EC is equal to
the squares of CF, FE ; therefore the rectangle
AD, DC, together with the square of EC, is
equal to the square of ED:, and CE is equal to EB; therefore the
rectangle AD, DC, together with the square of EB, is equal to the
square of ED : but the squares of EB, BD are equal to the square
(47. 1.) of ED, because EBD is a right angle; therefore the rectan-
gle AD, DC, together with the square of EB, is equal to the squares
of EB, BD : take away the common square of EB: therefore the re-
maining rectangle AD, DC is equal to the square of DB. Where-
fore, if from any point, &c. Q. E. D.

BOOK III.

THE ELEMENTS OF EUCLID.

Cor. If from any point without a circle,
there be drawn two straight lines cutting it,
as AB, AC, the rectangles contained by the
whole lines and the parts of them without the
circle, are equal to one another, viz. the rect-
angle BA, A£ to the rectangle CA, AF : for
each of them is. equal to the square of the
straight line AD which touches the circle.

PROP. XXXVn. THEOR.

Ir from a point without a circle there be drawn two straight
lines, one of which cuts the circle, and the other meets it ; if the
rectangle contained by the whole line, which cuts the circle, and
the part of it without the circle be equal to the square of the line
^hich meets it, the line which meets it shall touch the circle.*

Let any point D be taken without the chrcle ABC, and from it let
two straight lines DCA and DB be drawn, of which DCA cuts
the circle, and DB meets it, if the rectangle AD, DC be equal to the
square of DB ; DB touches the circle.

Draw (17. 3.) the straight line DE touching the circle ABC, find
its centre F, and join FE, FB, FD ; then FED is a right (18. 3.) an-
gle: and because DE touches the circle ABC, and DCA cuts It, the
rectangle AD, DC is equal (36. 3.) to the square of DE : but the
rectangle AD, DC is, by hypothesis, equal to the square of DB : there-
fore the square of DE is equal to the square of DB; and the straight
Ime DE equal to the straight line DB; D

and FE is equal to FB, wherefore DE, EF
are equal to DB, BF ; and the base FD is
common to the two triangles DEF, DBF ;
therefore the angle DEF is equal (8. 1.) to
the angle DBF : but DEF is a ri^t angle,
therefore also DBF is a right angle : and B
FB, if produced, is a diameter, and the
straight Ime which is drawn at right an-
gles to a diameter, firom the extremity of
it, touches (16. 3.) the circle : therefore
DB touches the circl? ABC. Wherefore,
if from a point, &c. Q. E. D.

• See Note^

THE

ELEMENTS OF EUCLID.

BOOK IV*

DEFINITIONS

A REicTiLiKeAi:. figure is said to be inscribed in another rectilineal
figure, when all the angles of the inscribed figure are upon the
sides of the figure in which it is inscribed, each
upon each.*

In like manner, a figure is said to be described about
another figure, when all the sides of the circum-
scribed figure pass through the angular points of
the figure about which it is ^escribed, each through

each.

A rectilineal figure is said to be inscribed in a
circle, when all the angles of the inscribed
figure are upon the circumference of the circle.

IV.

A rectilineal figure is said to be described about a circle, when eacb
side of the circumscribed figure touches the
circumference of the circle,

In like manner, a circle is said to be inscribed
in a rectilineal figure, when the circumference
of the circle touches each side of the figure.

« See Note.

BOOK IV.

THG fiLSMBinrt OF BUCLID.

»1

C -

VI.

A circle is said to be described about a rectili-
neal figure, when the circumiereiice of the
circle passes through all the angular points
erf* the figure about which it is described.

VIL

f A straight line is said to be placed in a circle, when the extremities
of it are in the circumference of the circle.

PROP. I. PROB.

Iif a given circle to place a straight line, equal to a given
straight line not greater than the diameter of the circle.

Let ABC be the given circle, and D the given straight line, not
greater than the diameter of the circle.

Draw BC the diameter of the circle ABC ; then, if BC be equal
to D, the thing required is done ; for in the drcle ABC a straight

line BC is placed equal to D ; but, if
it be not, BC is greater than D; make
CE equal (3. 1.) to D, and from the
centre C, at the distance C£^ describe
the circle AEF, and join C A : there-
fore, because C is the centre of the
circle AEF, CA is equal to CE; but
D is equal to CE; therefore D is equal p
to CA: wherefore in the circle ABC, a
straight line is placed equal to the

given straight line D, which is not greater than the diameter of the
circle. Which was to be done. ,

PROP. n. PROB.

Iir a given circle to inscribe a triangle equiangular to a given
triangle.

Let ABC be the given circle, and DEF the given triangle ; it is
required to inscribe in the circle ABC a triangle equiangular to the
triangle DEF.

Draw (17. 3.) the straight line GAH touching the circle in the point
A, and at the point A, in the straight Kne AH, make (23. 1.) the angle
HAC equal to the angle DEF ; and at the point A, in the straight Ihne
AG, make the angle GAB
eqiml to the angle DFE,
and join BC: therefore
because HAG touches the
circle ABC, and AC is
drawn from the point of
fM>ntact, the angle HAC is
equal (32. 3.) to the angle
ABC in the alternate seg-
ment of the circle : but

92 THB SLBMSHTS OP EUCLID. BOOK IT.

HAC is equal to the angle DEF, therefore also the angle ABC is
equal to D£1F ; for the same reason, the angle ACB is equal to the
angle JDFE; therefore the remaining angle BAG is equal (32. 1.) to
the remaining angle EDF : wherefore the triangle ABC is equiangu-
lar to the triangle DEF, and it is inscribed in the circle ABC. Which
was to be done.

PROP. m. PROR

About a ^iven circle to describe a triangle equiangular to a
given triangle.

Let ABC be the given circle, and DEF the given triangle ; it is
required to describe a triangle about the circle ABC equiangular to
the triangle DEF.

Produce EF both ways to the points G, H, and find the centre K
of the circle ABC, and from it draw any straight line KB ; at the
poi^t K in the straight line KB, make (23. 1.) the angle BKA equal
to the angle DEG, and the angle BKC equal to the angle DFH ; and
through the points A, B, C, draw the straight lines LAM, M BN, NCL
touching (17. 3.) the circle ABC : therefore because LM, MN, NL
touch the circle ABC in the points A, B, C, to which from the centre
are drawn KA, KB, KC, the angles at the points A, B, C are right
(18. 3.) angles: and because the four angles of the quadrilateral
figure AMBK are equal to four right angles, for it can be divided
into two triangles: and that two of them KAM, KBM are right
angles, the other two AKB, AMB are equal to two right angles : but
the angles DEG, DEF L

are likewise equal (13.
L) to two right an-
gles; therefore the an-
gles AKB, AMB are
equal to the angles
DEG, DEF of which

AKB is equal to DEG; ,. i x i; v h

wherefore the remain- /\ / x ^ r H

ing angle AMB is
equal to the remain-
ing angle DEF; in M B . N
like manner, the angle LNM may be demonstrated to be equal to
DFE : and therefore the remaining angle MLN is equal (32. 1.) to
the remaining angle EDF : wherefore the triangle LMN is equiangu-
lar to the triangle DEF : and it is described about the circle ABC.
Which was to be done.

PROP. IV. PROB.

To inscribe a circle in a given triangle.*

Let the given triangle be ABC ; it is required to inscribe a circle
in ABC.

Bisect (9. 1.) the angles ABC, BCA by the straight lines BD,

» See Jf 6te.

BOOK lY.

THE ILfiMfiMTS OF BCOC«ID.

CD meeting one another at the point D, from which draw (12. 1.)
DE, DF, DG perpendiculars to AB, A

BC, CA : and because the angle EBD
is equal to the angle FBD, for the
angle ABC is bisected by BD, and that
the right angle BED is equal to the
right angle BFD, the two triangles
£BD, FBD have two angles of the
one equal to two angles pf the other^
and the side BD, which is opposite to
one of the equal angles in each is com-
mon to both; therefore their other
sides shall be equal (26. 1.) ; where-
fore DE is equal to DF : for the same reason, DG is equal to DF :
therefore the three straight lines DE, DF, DG are equal to one ano-
ther, and the circle described from the centre D, at the distance of
any of them, shall x)ass through the extremities of the other two,
and touch the straight lines, AB BC, CA, because the angles at the
points E, F, G are right angles, and the straight line which is drawn
from the extremity of a diameter at right angles to it, touches (16. 3.)
the circle : therefore the straight lines AB, BC, CA do each of them
touch the circle, and the circle EFG is inscribed in the triangle ABC
Which was to be done.

PROP, V. PROB,
To describe a circle about a given triangle.*

Let the given triangle be ABC ; it is required to describe a circle
about ABC.

Bisect (10. 1.) AB, AC to the points D, E, and from these points
draw DF, EF at right angles (11. 1.) to AB, AC ; DF, EF produced
A A A

C B

meet one another : for, if they do not meet, they are parallel, where-
fore AB, AC, which are at right angles to them, are parallel ; which
Iff absurd : let them meet in F, and join FA ; also, if the point F be
not in BC, join BF, CF ; then, because AD is equaj to DB, and DF
common, and at right angles to AB, the base AF is equal (4. 1.) to
the base FB : in like manner, it may be shown, that CF is equal to
FA ; and therefore BF is equal to FC ; and FA, FB, FC arc equal
to one another : wherefore the circle described from the centre F,
at the distance of one of them, shall pass through the extremities of

See Note.>

84 TRfi ELEMENTS OP EUCLID. BOOK IT.

the other two, and be described about the triangle ABG. Which
was to be done.

Cor. And it is manifest, that when the centre of the circle falls
within the triangle, each of its angles is less than a right angle, eadi
of them being in a segment greater than a semicircle ; but, When
the centre is in one of the sides of the triangle, the angle opposite to
this side, being a semicircle, is a light angle; and, if the centre &lls
without the triangle, the angle opposite to the side beyond which it
is, being in a segment less than a senncircle, is greater than a right
angle : wherefore, if the given triangte be acute angled, the centre
of the circle fells within it ; if it be a right angled triangle, the centre
is in the side opposite to the right angle ; and if it be an obtuse an-
gled triangle, the centre falls without the triangle, beyond the side
opposite to the obtuse angle.

PROP. V. PROB.

To inscribe a square in a given drcle.

Let ABC be the given circle ; it is requred to inscribe a square in
ABCD.

Draw the diameters AC, BD at right angles to one another ; and
join AB, BC, CD, DA ; because BE is equal to £33, for E is the
centre, and that EA is common, and at A

right angles to BD ; the base BA is equal
(4. 1.) to the base AD; and for the same
reason, BC, CD are each of them equal to

BA, or AD ; therefore the quadrilateral fi- g IX ^ \ - ^

gure ABCD is equilateral. It is also rect- "^^ ^ -

angular ; for the straight line BD, being
the diameter of the circle ABCD, BAD is a
semicircle ; wherefore the angle BAD is a
right (31. 3.) angle ; for the same reason C

each of the angles ABC, BCD, CDA is a right angle; therefore the
quadrilateral figure ABCD is rectangular, and it has been shown to
be equilateral ; therefore it is a square ; and it is inscribed^in the
circle ABCD. Which was to be done.

PROP. Vn. PROB.

To describe a square about a given circle.

Let ABCD be the giv^i circle : it is required to describe a square
•about it

Draw two diameters AC, BD, of the circle ABCD, at right
.angles to one anoth^, and through the points A, B, C, D» draw
(l?. 3.) FG, GH, HK, KF touching the curcle; and because FG
touches the circle ABCD, and EA is drawn from the centre £ to
the point of contact A, the angles at A nre right (18. 3.) angles;
for the same reason, the angles at the points B, C, D, are right

BOOK IV.

THfi ELBMSNT8 OF EUCLID.

angles; and )because the angle AEB is
a right angle, as likewise is EBG, GH
is parallel (28. 1.) to AC ; for the same
reason, AC is parallel to FK, and in like'
manner GF, HK may each of them be de-
monstrated to be parallel to BED ; there-
fore the figures GK, GC, AK, FB, BK are
parallelograms ; and GF is therefore equal
(34. 1.) to HK, and GH to FK ; and be-
cause AC is equal to BD, and that AC H C K
is equal to each of the two GH, FK; and BD to each of the two
GF^ HK : GH, FK are each of them equal to GF or HK ; therefore
the quadrilateral figure FGHK is equilateral. It is also rectangular :
for GBEA being a parallelogram, and AEB a right angle, AGB (34.
1.) is likewise a right angle : in the same manner, it may be shown
that the angles at H, K, F are right angles ; therefore the quadH-
lat^al figure FGHK is rectangular, and it was demonstrated to be
equilateral : therefore it is a square ; and it was described about the
circle ABCD. Which was to be done.

PROP. VIL PROB.
To inscribe a circle in a given square.

Let ABCD be the gi\&en square j it is required to inscribe a circle
in ABCD.

Hsect (10. 1.) each of the sides AB, AD, in the points F, E, and
through E draw (31. 1.) EH parallel to AB or DC, and through F
draw FK parallel to AD or BC ; therefore each of the figures AK,
KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite
sides are equal (34. 1.) and because AD is equal to AB, and that AE
is the half of AD, and AF the half of AB, EA is equal to AF ; where-
fore the sides opposite to these are equal, A E D
viz. FG to GE: in the same manner, it
may be demonstrated that GH, GK are
each of them equal to FG or GE; there-
fore the four straight lines GE, GF, GH,
GK, are equal to one another; and the F
circle described from the centre G, at the
distance of one of them, shall pass thro^igh
the extremities of the other three, and touch
the straight lines AB, BC, CD, DA; be- B
cause the angles at the points E, F, H, K are right (2§. 1») angles,
and that the straight line which is drawn firom the extremity of a
diameter, at right angles to it, touches the circle (16. 3.) therefore
each of the straight lines AB, BC, CD, DA touches the circle, whidi
therefore is inscribed in the square ABCD. Which was to be done

PROP. IX. PROB,

To describe a given square.

Let ABCD be the given square ; it is required tp describe a circle
about it.

THB SLEMENTS OJP JfEUCLID.

BOOK iV*

Join AC, BD cutting o&e another in E ; and because DA is equal
to AB, and AC common to the triangles DAC, BAC, the two sides
DA, AC are equal to the two BA, AC ; and
the base DC is equal to the base BC ; where-
fore the angle DAC is equal (8. 1.) to the
angle BAC, and the angle DAB is bisected by
the straight line AC : in the same manner,
it may be demonstrated that the angles ABC,
BCD, CDA are severally bisected by the
straight lines BD, AC ; therefore, t)ecause the
angle DAB is equal to the angle ABC, and
that the angle EAB is the half of DAB, andEBA the half of ABC;
the angle EAB is equal to the angle EBA ; wherefore the side EA.
is equal (6. 1.) to the side EB: in the same manner, it may be de-
monstrated that the straight lines EC, ED are each of them' equal to
EA or EB : therefore the four straight lines EA, EB, EC, ED are
equal to one another; and the circle described from the centre E,
at the distance of one of them, shall pass through the extremities of
the other three, and be described about the square ABCD. Which
was to be done.

PROP. X. PROB.

To describe an isosceles triangle, having each of the angles
at the base double of the third angle.

Take any straight line AB, and divide (11. 2.) it in the point C, so
that the rectangle AB, BC be equal to the square of CA ; and from
the centre A, at the distance AB, describe the circle BDE^ in which
place (1. 4.) the straight line BD is equal to AC, which is not greater
than the diameter of the circle BDE ; join DA, DC, and about the .
triangle ADC describe (5. 4.) the circle ACD ; the triangle ABD i»
such as is required, that is, each of the angles ABD, ADB is double of
the angle BAD. .

Because the rectangle AB, BC is equal to the square of AC, ana
that AC is equal to BD, the rectangle AB, BC is equal to the square
of BD ; and because from the point
B without the circle ACD two
straight lines BCA, BD are drawn
to the circumference, one of which
cuts, and the other meets the
circle, and that the rectangle AB,
BC contained by the whole . of
the cutting line, and the part of
it without the circle is equal to
the square of BD which meets
it ; the straight line BD touches
(37. 3.) the circle ACD; and
because BD touches the circle,
and DC is drawn from the point
of contact Di, the angle BE>C is
equal (32. 3.) to the angle DAC

BOOK IV.

THl ELEMENTS OF EUCLID.

in the alternate segment of the circle; to each of these add the angle
CDA: therefore the whole angle BDA is equal to the two angles
CDA, DAC ; but the exterior angle BCD is equal (32. 1.) to the an-
gles CDA, DAC ; therefore also BDA is equal to BCD ; but BDA is
equal (5. 1.) to the angle CBD, because the side AD is equal to the
side AB ; therefore CBD, or DBA is equal to BCD ; and consequent-
ly the three angles BDA, DBA, BCD are equal to one anottier ; and
because the angle BDC is equal to the angle BOD, the side BD is
equal (6. 1.) to the side DC ; but BD was made equal to CA ; there-
fore also CA is equal to CD, and the angle CDA equal (5. I.) to the
angle DAC; therefore the angles CDA, DAC together, are double of
the angle DAC : but BCD is equal to the angles CDA, DAC ; there-
fore also BCD is double of DAC„ and BCD is equal to each of the
angles BDA, DBA; each therefore of the angles BDA, DBA, is double
of the angle DAB; wherefore an isosceles triangle ABD is described,
having each of the angles at the base double of the third angle.
Which was to be done. •

PROP. XI. PROP.

To inscribe an equilateral and equiangular pentagon in a given
circle.

Let ABODE be the given cu-cle ; it is reqpired to inscribe an equi-
lateral and equiangular pentagon in the circle ABODE.

Describe (10. 4.) an isosceles triangle FGH, having each of the
angles at G, H, double of the angle at F ; and in the circle ABODE
inscribe (2. 4.) the triangle AOD equiangular to the triangle FGH,
so that the angle CAD be equal
to the angle at F,fand each of the
angles ACD, CDA equal to the
angle at G or H ; wherefore each
of the angles ACD, CDA is double
of the angle CAD. Bisect (9. 1.)
the angles ACD, CDA by the
straight Imes CE, DB : and jom
AB, BO, DE, EA. ABODE is the
pentagon required.

Because each of the angles

ACD, CDA is double of CAD, ^
and are bisected by the straight Imes CE, DB, the five angles DA C

ACE, EOD, CDB, BDA are equal to one another ; but equal angles
stand upon equal (26. 3.) circumferences ; therefore the five circum-
ferences AB,.BC, CD, DE, EA are equal to one another: and equal
circumferences are subtended by equal (29. 3.) "straight lines; there-
fore the five straight lines AB, BO, CD, DE, EA are equal to one
another. Wherefore the pentagon ABODE is equilateral. It is also
equiangiilar ; because the circumference AB is equal to the circum«
ference DE : if to^each be added BOD, the whole ABOD is equal to
the whole EDCB: and the ang)e AED stands, on the circumference
ABOD, and the angle BAE on Ihe circumference EDCB ; therefore

THE IBLBKBin'S OF EUGUD.

BOOS IV.

the angle BAE is equal (27. 3.) to the angle AED : for the same rea-
son, each of the angles ABC» BCD, ODE is equal to the angle BAE,
or AED : therefore the pentagon ABCDE is equiangular. ; and it has
been shown that it is equilateral. Wherefore, in the given circle^ an
equilateral and equiangular pentagon has been inscribed. Which
was to be done.

PROP. XII. PROB. '

To describe an equilateral and equiangular pentagon about a
given circle.

Let ABCDE be the given circle ; it is required to deseribe an equi-
lateral and equiangular pentagon about the circle ABCDE.

Let the angles of a pentagon, inscribed in the circle, by the last
proposition, be in the points A, B, C, D, E, so that the circumferences
AB, BC, CD, DE, EA are equal (11.4.); and through the points A,
B, C, D, E draw GH, HK, KL, LM, MG, touching (17. 3.) the circle;
take the centre P, and join PB, FK, PC, PL, PD : and because the
straight hne BIL touches the circle ABCDE in the point C, to which
PC is drawn from the centre P, PC is perpendicular (18. 3.) to KL;
therefore each of the angles at C is a right angle : for the same rea-
son, the angles at the points B, D are right angles : and because FCK
is a right angle, the square of PK is equal (47. 1.) to the squares of
PC, CK : for the same reason, the square of FK is equal to the squares
of PB, BK: therefore the squares of PC, CK are equal to the squares
of FB, BK, of which the square of PC is equal to the square of PB;
the remaining square of CK is therefore equal to the remaining
square of BK, and the straight line CK equal to BK: and because
PB is equal to PC, and PK common to the triangles BFK, CFK,
the two BP, PK are equal to the two CF, PK ; and the base BK
is equal to the base KC ; therefore the angle BPK is equal (8, i.) to
the angle KFC, and the angle BKP to FKC j wherefore the angle
BPC is double of the angle KPC, and mCC double of FKC ; for
the same reason, the angle CPD is double of the angle CPL, and
CLD double of CLP : and because the circumference BC is equal
to the circumference CD, the angle BPC is equal ^27. 3.) to the
angle CPD ; and BPC is double of the
angle KPC, and CPD double jof CPL;
therefore the angle KPC is equal to
the angle CPL ; and the right angle
FCK is equal to the right angle
FCL : therefore, in the two triangles
FKC, PLC, there are two angles of
one equal to two angles of the other,
each to each, and the side PC, which
is adjacent to the equal angles in
each, is common to both ; therefore

the other sides shall be equal (26. K C L

1.) to the other sides, and the third an^e to the third angle: there-
fore the straight line KC is equal to CL, and the angle FKC to the

BOOK IV.

THB ELEMENTS OF EUCLID.

angle FLC : and because KC is equal to CL, KL is double of KC :
in ^e same manner, it may be shown that HK is double of BK :
and because BK is equal to KC, as was demonstrated, and that KL
is double of KC, and HK double of BK, HK shall be equal to KL :
in like manner it may be shown that GH, GM» ML are each of them
equal to HK or KL: therefore the pentagon GHKLM is equUateral.
It is also equiangular ; for, since the angle FKC is equal to the angle
FLO, and that the angle BKL is double of the angle FKC, and KLM
double of FLC, as was before demonstrated, the angle HKL is equal
to KLM : and in like manner it may be shown, that each of the
angles KHG, HGM, GML is equal to the angle HKL or KLM : there-
fore the five angles GHK, HKL, KLM, LMG, MGH being equal to
one another, the pentagon GHKLM is equiangular : and it is equila-
teral, as was demonstrated; and it is described about the circle
ABODE. Whidi was to be done.

PROP. XIII. PROB.

To inscribe a circle in a given equilateral and equiangular
pentagon.

Let ABODE be the given equilateral and equiangular pentagon: it
is required to inscribe a circle in the pentagon ABCDE.

Bisect (9. 1.) the angles BCD, CDE by the straight lines CF, DP,,
and from the point F, in which they meet, draw the straight lines
PB, FA, FE ; therefore, since BC is equal to CD, and CF common to
the triangles BCF, DCF, the two sides BC, CF, are equal to the two
DC, CF ; and the angle BCF is equal to the angle DCF ; therefore
the base BF is equal (4. 1) to the base FD, and the other angles to
the other angles, to which the equal sides are opposite ; therefore
the angle CBF is equal to the angle CDF : and because the angle
CDE is double of CDF, and that CDE is equal to CBA, and CDF to
CBF ; CBA is also double of the angle ^

CBF ; therefore the anglp ABF is equal
to the angle CBF ; wherefore the angle
ABC is bisected by the straight line BF:
in the same manner it may be demon-
strated, that the angles BAE, AED are ^
bisected by the straight lines AF, FE :
from the point F draw (12. 1.) FG, FH^ jj
FK, FL, FM, perpendiculars to the
straight lines AB, BC, CD, DE, EA;
and because the angle HCF is equal to
KCF^ and the right angle FHC equal to
the right angle FKC; in the triangles
FHC, FKC there are two angles of one equal to two angles of lAe
other, and the side FC, which is opposite to one of the equal angle»
in each, is^ common to bpth ; therefore the other sides shall be equal
(26. 1.) each to each ; wherefore the perpendicular FH is equal to
the perpendicular FK: m the same maimer it may be demonstrated

90 THE ELEMENTS OP EUCLID. BOOK lY.

that PL, FM, PG are each of them equal to FH, or FK ; therefore
the five straight lines FG, FH, FK, FL, FM are equal to one another:
wherefore the circle described from the centre P, at the distance of
one of these five, shall pass through the extremities of the other four,
and touch the straight lines AB, BC, CD, DE, EA, because the angles
at the points G, H, K, L, M are right angles ; and that a straight line
drawn from the extremity of the diameter of a circle at right angles
to it, touches (16. 3,) the circle: therefore each of the straight lines
AB, BC, CD, DE, EA touches the circle ; wherefore it is inscribed in
the pentagon ABCDE. Which was to be done,

PROP. XIV. PROB.

To describe a circle about a given equilateral and equiangular
pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it
is required to describe a circle about it.

Bisect (9. L) the angles BCD, CDE by the straight lines CF, FD,
and from the point F, in which they meet, draw the straight lines
FB, FA, FE, to the points B, A, R It A

may be demonstrated, in the same manner
as in the precedmg proposition, that the an-
gles CBA, BAE, AED, are bisected by the B
straight lines FB, FA, FE: and because the
angle BCD is equal to the angle CDE, and
that PCD is the half of the angle BCD, and
CDF the half of CDE ; the angle PCD is
equal to FDC : wherefore the side CP is
equal (6. 1.) to the side FD : in like manner
it may be denionstrated that FB, FA, FE
are . each of them equal to PC or FD : therefore the five straight
lines FA, FB, PC, FD, FE are equal to one another; and the circle
described from the centre F, at the distance of one of them, shall
pass {through the extremities of the other four, and be described
, about the equilateral and equiangular pentagon ABCDE. Which
was to be done.

PROP. XV. PROB.

To inscribe an equilateral and equiangular hexagon in a given
circle.*

Let ABCDEP be the given circle ; it id required to inscribe an
equilateral and equiangular hexagon in it.

Find the centre G of the circle ABCDEP, and draw the dia-
meter AGD ; and from D as a centre, at the distance DG, de-
scribe the circle EGCH, join EG, CG, and produce them to the

* See Note.

BOOK IV.

THB GLBHENTS OP EUCLID.

IK>i]its B, F ; and join A6» BC, CD, DE, EF, FA : the hexagon
ABCDEF, is equilateral and equiangular.

Because G is the centre of the circle ABCDEF, GE is equal to
GD : and because D is the centre of the circle EGCH, DE is equal
to DG : wherefore GE is equal to ED, and the triangle EGD is equi-
lateral ; and therefore its three angles EGD, GDE, DEG are equal
to one another, because the angles at the base of an isosceles trisui-
gle are equal (5. 1.) ; and the three angles of a triangle are equal
(32. 1.) to two right angles; therefore the angle EGD is the third
part of two right angles : in the same manner, it may be demon-
strated, that the angle DGC is also the third A
part of two right angles : and because the
straight line GC makes with EB the adja-
cent angles EGC, CGB equal (13. 1.) to two
right angles; the remaining angle CGB is
the third part of two right angles ; therefore
the angles EDG, DGC, CGB are equal to
one another : and to these are equal (15. 1.) £
the vertical opposite angles BGA, AGF,
FGE : therefore the six angles EGD, DGC,
CGB, BGA, AGF, FGE are equal to one
another: but equal angles stand upon equal
(26. 3.) circunrferences ; therefore the six
circumferences AB, BC, CD, DE, EF, FA
are equal to one another : and equal circum-
ferences are subtended by equal (29. 3.) straight lines; therefore
the six straight lines are equal to one another, and the hexagon
ABCDEF, is equilateral. It is also equiangular ; for, since the cir-
cumference AE is equal to ED, to each of these add the circumfe-
rence ABCD : therefore the whole circumference FABCD shall be
equal to the whole EDCBA : and the angle FED stands upon the
circumference FABCD, and the angle AFE upon EDCBA ; there-
fore the angle AFE is equal to FED : in the same manner it may be
demonstrated that the other angles of the hexagon ABCDEF are
each of them equal to the angle AFE or FED ; therefore the hexa-
gon is equiangular ; and it is equilateral, as was shown ; and it is
inscribed in the given circle ABCDEF. Which was to be done.

Cor. From this it is manifest, that the side of the hexagon is
equal to the straight line from the centre, that is^ to the semidiameter
of the circle.

And if through the points A, B, C, D, E, F there be drawn straight
lines touching the circle, an equilateral and equiangular hexagon
shall (be described about it, which may be demonstrated from what
has been said of the pentagon; and likewise a circle may be in-
scribed in a given equilateral and equiangular hexagon, and circum-
scribed about it, by a method like to that used for the pentagon.

03 THE ELfiMKNTS OP EUCLID. BOOK IV.

PROP. XVI. PROB.

To inscribe an equilateral and equiangular quindecagon, in a
given circle.*

Let ABCD be the given circle ; it is required to inscribe an equi-
lateral and equiangular quindecagon in the circle ABCD.

L.et AC be the side of an equilateral triangle inscribed (2* 4.) in
the circle, and AB the side of an equilateral and equiangular penta-
gon inscribed (11. 4.) in the same; therefore, if such equal parts as
the whole circumference ABCDF contains fifteen, the circumference
ABC, being the third part of the whole, A

contains five; and the circumference AB,
which is the fifth part of the whole, con-
tains three ; therefore BC their difference
contains two of the same parts : bisect
(11. 4.) BC in E; therefore BE, EC are,
each of them, the fifteenth part of the E|
whole circumference ABCD: therefore, if
the straight lines BE, EC be drawn, and
straight lines equal to them be placed (1.

4.) around in the whole circle, an equila-

teral and equiangular quindecagon shall be inscribed in it. Which
was to be done.

And in the same manner as was done in the pentagon, if through
the points of division made by inscribing the quindecagon, straight
lines be drawn touching the circle, an equilateral and equiangular
quindecagon shall be described about it: and likewise as in the
pentagon, a circle may be inscribed in a given equilateral and equi-
angular quindecagon, and circumscribed about it.

* See Note.

THE

ELEMENTS OF EUCLID.

BOOK V.

DEFINITIONS.

A LESS magnitude is said to be a part of a greater magnitude, when
the less measures the greater, that is, * when the less is contained
a certain number of times exactly in the greater.*

A greater magnitude is said to be a multiple of a less, when the
greater fs measured by the less, that is, * when the greater contains
the less a certain number of times exactly.'

ni.

'Ratio is a mutual relation of two magnitudes of the same kind to
one another, in respect of quantity.' *

IV.

Magnitudes are said to have a ratio to one another, when the less
can be multiplied so as to exceed the other.

The first of four magnitudes, is said to have the same ratio to the
second, which the third has to the fourth, when any equimultiples
whatsoever of the first and third bemg taken, and any equimulti-
ples whatsoever of the second and fourth ; if the multiple of the
first be less than that of the second, the multiple of the third is
also less than that of the fourth ; or, if the multiple of the first be
equal to that of the second, the multi^de of the third is also equal
to that of the fourth ; or, if the multiple of the first be greater
than that of the second, the multiple of the third is also greater
than that of the fourth.

VI.

Magnitudes which [have the same ratio are called proportionals.

« See Note.

94 THE GLEMENT8 OF EUCLID. ' BOOK V.

N. B. * When four magnitudes are proportionals, it is usually ex-
pressed by saying, th'e first is to the second, as the third to the
fourth.'

VII.

When of the equimultiples of four magnitudes (taken as in the fifth
definition) the multiple of the first is greater than that of the
second, but the multiple of the third is not greater than the multi-
ple of the fourth ; then the first is said to have to the second a
greater ratio than the third magnitude has to the fourth ; and, on
the contrary, the third is said to have to the fourth a less ratio
than the first has to the second.

vm.

♦Analogy, or proportion, is the similitude of ratios.'

IX.
Proportion consists in three terms at least.

X. '

When three magnitudes are proportionals, the first is said to have
to the third the duplicate ratio of that which it has to the second.

When four magnitudes are continual proportionals, the first is said
to have to the fourth the triplicate ratio of that which it has to
the second, and so on, quadruplicate, &c. increasing the denomi-
nation still by unity, in any number of proportionals.

Definition A, to wit, of compound ratio.

When there are any number of magnitudes of the same kind, the first
is said to have to the last of them the ratio compounded of the
ratio which the first has to the second, and of the ratio which the
second has to the third, and of the ratio which the third has to the
fourth, and so on unto the last magnitude. .

For example, if A, B, C, D be four magnitudes of the same kind, the
first A is said to have to the last D the ratio compounded of the
ratio of A to B, and of the ratio B to C, and of the ratio of C to
D ; or, the ratio of A to D is said to be compounded of the ratios
of A to B, B to C, and C to D.

And if A have to B the same ratio which E has to F ; and B to C,
the same ratio that G has to H; and C to D, the same that K
has to L ; then, by this definition, A is said to have to D the ratio
compounded of ratios which are the same with the ratios of E to
F, G to H, and K to L: and the same thing is to be understood
wh^n it is more briefly expressed, by saying A has to D the ratio
compounded of the ratios of E to F, G to Hi and K to L.

In like planner the same things being supposed, if M have to N the
same ratio which A has to D: then for shortness' sake, M
is said to have to N, the ratio compounded of the ratios of E
to F, G to H, and K to L.

XII.

In proportionals, the antecedent terms are called homologous to
one another, as also the consequents to one another.

BOOK V. THB £LCM£NTfl OF EUCLID. 95

* Gveometers make use of the Mowing technical words to sigi^ cer-
tain ways of changing either the order or magnitude of propor-
tionals, so as that they continue still to be proportionals.'

XIIL

Permutando, or alternando, by permutation, or alternately ; this word
is used when there are four proportionals, and it is inferred, that
the first has the same ratio to the third, which the second has to
the fourth ; or that the first is to the third, as the second to the
fourth; as is shown in the 16th prop, of this 5th book.^

XIV.

Invertendo, by inversion ; when there are four proportionals, and it
, is inferred, that the second is to the first, as the fourth to the third.
Prop. B. book 5.

XV.

Componendo, by composition; when there are four proportionals,
and it is inferred, that the first, together with the second, is to the
second, as the third, together with the fourth, is to the fourth.
18th Prop, book 5.

XVI.

Dividendo, by division ; when there are four proportionals, and it is
inferred, that the excess of the first above the second, is to the
second, as the excess of the third above the fourth, is to the fourth.
17th Prop, book 5.

xvn.

Convertendo, by conversion ; when there are four proportionals, and
it is inferred, that the first is to its excess above tlie second, as
the third to its excess above the fourth. Prop. E. book 5.

xvm.

£x aequali (sc. distantia), or ex aequo, from equality of distance when
there is any number of magnitudes more than two, and as many
others, so that they are proportionals when taken two and two of
each rank, and it is inferred, that the first is to the last of the first
rank of magnitudes, as the first is to the last of the others : * Of
this there are the two following kinds, which arise from the diffe-
rent order in which the magnitudes are takeit two and two.'

XIX.

Ex aequsdi, from equality; this term is used simply by itself, when
the first magnitude is to the second of the first rank, as the first

* See Note.

THfi ELBMfiNTS OP EUCLID. BOOK V.

to the second of the other ronk: and as the second is to the tburd
of the first rank, so is the seccmd to the third of the other : and so
on in order, and the inference is as mentioned in the preceding
definition ; whence this is called ordinate proportion. It is demon-
strated in 22d Prop, book 5.

XX.

£x sequali, in proportione perturbata, seu inordinata ; from eqaality,
in perturbata or disorderly proportion ;* this term is used when
the first magnitude is to the second of the first rank, as the last
but one is to the last of the second rank ; and as the second is to
the third of the first rank, so is the last but two to the last but one
of the second rank ; and as the third is to the fourth of the first
rank, so is the third from the leu^t to the last but two of the second
rank : and so on in a cross order : and the inference is as in the
18th definition. It is demonstrated in the 23d Prop, of book 5.

AXIOMS.

Equimultiples of the same, or of equal magnitudes, are equal to one
another.

II.

Those magnitudes of which the same, or equal magnitudes are equi-
multiples, are equal to one another.

m. .

A multiple of a greater magnitude is greater than the same multiple
of a less.

IV.

That magnitude of which a multiple is greater than the same multi-
ple of another, is greater than that other magnitude.

PROP. L THEOR.

If any number of magnitudes be equimultiples of as many,
each of each ; what multiple soever any one of them is of its
part, the same multiple shall all the first magnitudes be of aU the
other.

Let any number of magnitudes AB, CD be equimultfples of as
many others, E, F, each of each ; whatsoever multiple AB is of E,
the same multiple shall AB and CD together be of E and F together.

Because AB is the same multiple of £ that CD is of F, as many
magnitudes as are in AB equal to E, so many are there in CD

*4Prop.lib.3. Arcbiiaedii de spluBn tt cylindro

BOOK V.

THE KLBMBNTS OF fiUCUD.

G—

equal to F< Divide AB into magnitudes equal
to E, viz. AG, GB ; and CD into CH, HD equal
each of them to F : the number therefore of t}ie
magnitudes CH, HD shall be equal to the num-
ber of the others, AG, GB ; and because AG is
equal to E, and CH to F, therefore AG and CH
together are equal to (Ax. 2. 5.) E and F to-
gether: for the same reason, because GB is
equal to £, and HD to F ; GB and HD together
are equal to E and F together. Wherefore, as
many magnitudes as are in AB equal to E, so
many are there in AB, CD together equal to E
and F together. Therefore, whatsoever multi-
pie AB is of E, the same multiple is AB and CD
together of E and F together.

Therefore, if any magnitudes, how many soever, be equimultiples
of as many, each of each, whatsoever multiple any one of them is
of its part, the same multitple shall all the first magnitudes be of all
the other : * For the same demonstration holds in any number of
* magnitudes which was here applied to two.' Q,. E. D.

H—

PROP. n. THEOR.

If the first magnitude be the same multiple of the second that
the third is of the fourth, and the fifth the same multiple of the
second that the sixth is of the fourth ; then shall the first together
ivith the fifth be the same multiple of the second, that the third
together with the sixth is of the fourth.

Let AB the first, be the same multiple of C the second, that
DE the third is of F the fourth ; and BG the fifth, the same mul-
tiple of C the second, that EH the
sixth is of F the fourth : then is AG A D

the first, together with the fifth, the
same multiple of C the second, that
DH the third, together with the sixths
is of F the fourth.

Because AB is the same multiple
of C, that DE is of F ; there are as
many magnitudes in AB equal to
C, as there are in DE equal to F : G
in like manner, as many as there

are in BG equal to C, so many are there in EH equal to F : as
many> then, as are in the whole AG equal to C, so many are there
in the whole DH equal to F : thearefore AG is the same multiple

B—

TH£ ELEMBNTS OP EUCLID.'

SOOK V.

Of C, that DH is of F; that is, AG the
first and fifth together, is the same multi-
ple of the second C, that DH the third
and sixth together, is of the fourth of F.
If, therefore, the first be the same multi-
ple, &c. Q,. K D.

Cor. *From this it is plain, that, if

• any number of magnitudes AB, BG,

* GH, be multiples of another C ; and as

• many DE, EK, KL, be the same multi-

* pies of F, each of each ; the whole of

* the first, viz. AH, is the same multiple of

• C, that the whole of the last, viz. DL, is
*ofF.'

B-

G—

E—

L F

PROP. m. THEOR.

Ir the first be the same multiple of the second, which the third
is of the fourth ; and if of the first and third there be taken equi-
multiples, these shall be equimultiples, tlie one of the second, and
the other of the fourth.

K-

L-^

Let A the first, be the same multiple of B the second, that C the
third is of D the fourth ; and of A, C let the equimultiples EF, GH
be taken ; then EF is the same multiple of B that GH is of D.

Because EF is the same multiple of A that GH is of C, there are
as many magnitudes in EF equal to A ; as are in GH equal to : let
EF be divided into the magnitudes
£K, EF, each equal to A, and GH
into GL, LH, each equal to C : the
number therefore of the magnitudes
£E, EF, shall be equal to the num-
ber of the others GL, LH: anti be-
cause A is the same multiple of B,
that C is of D, and that EE is
equal to A, and GL to C ; there-
fore EE is the same multiple of B,
that GL is of D ; for the same
reason EF is the same multiple of
B, that LH is of D ; and so, if
there be more parts in EF, GH
equal to A, C : because, therefore,
the first EE is the same multiple of the second B, which the third
GL is of the fourth D, and that the fifth EF is the same multiple of
the second B, which the sixth LH is of the fourth D ; EF the first
together with the fifth, is the same multiple (2. 5.) of the second B,
which GH the third, together with the sixth, is of the fourth D. If,
therefore, the first, &c. Q,. E. D.

E A B

I
G

C D

BOOK T.

THC ELEMENTS OF EUCLID.

PROP. IV. THEOR.

If the first of four magnitudes has the same ratio to the second
which the third has to the fourth, then any equimultiples what-
ever of the first and third shall have the same ratio to any equi-
multiples of the second and fourth, viz. * the equimultiple of the

* first shall have the same ratio to that of the second, which the

* equimultiple of the third has to that of the fourth.'*

Let A the first, have to B the second, the same ratio which the
third C has to the fourth D ; and of A and C let there be taken any
equimultiples whatever E, F : and of B and D any equimultiples
whatever G, H : then E has the same
ratio to G, which F has to H.

Take of E and F any equimulti-
ples whatever K, L, and of G, H,
any equimultiples whatever M, N:
then, because E is the same multiple
of A, that F is of C ; and of E and
F have been taken equimultiples K,
L; therefore K is the same multiple
of A, that L is of C (3. 5.) ; for the
same reason, M is the same multiple
of B, that N is of D: and because,
as A is to B, so is C to D (Hypoth.)
and of A and C have been taken cer-
tain equimultiples K, L; and of B
and D have been taken certain equi-
multiples M, N ; if therefore K be
greater than M, L is greater than N ;
and if equal, equal ; if less, less (5.
def. 5.). And K, L are any equimul-
tiples whatever of E, F ; and M, N
any whatever of G, H : as therefore
E is to G, so is (5. def. 5.) F to H.
Therefore, if the first, &c. Q. E. D.

Cor, Likewise, if the first have the
same ratio to the second, which the
third has to the fourth, then also any
equimultiples whatever of the first and third have the same ratio to
the second and fourth: and in like manner, the first and the third
have the same ratio to any equimultiples whatever of the second and
fourth.

Let' A the first, have to B the second, the same ratio which the
third C lias to the fourth D, and of A and C let E and F be any
equimultiples whatever ; then E is to B, as F to D.

Take of E, F any equimultiples whatever K, L, and of B, D any
equimultiples whatever G, H; then it may be demonstrated, as be-

K E A B G M

L F C D H N

« See Note.

100

THE ELEMENTS OF EUCLID.

BOOK V.

fore, that K Is the same multiple of A, that L is of C : and because
A is to B, as C is to D, and of A and C certain equimultiples have
been taken, viz. K and L ; and of B and D certain equimidtiples G,
H ; therefore if K be greater than G, L is greater than H ; and if equal,
equal ; if less, less (5. def. 5.) : and K, L are any equimultiples of E,
F, and G, H any whatever of B, D : as therefore E is to B, so is F to
D : and in the same way is the other case demonstrated.

PROP. V. THEOR.

If one magnitude be the same multiple of another, which a
magnitude taken from the first is of a magnitude taken from the
other; the remainder shall be the same multiple of the remainder,
that the whole is of the whole.

A—

Let the magnitude AB be the same multiple
of CD, that AE taken from the first, is of CF
taken from the other; the remainder EB shall
be the same multiple of the remainder FD, that
the whole AB is of the whole CD.

Take AG the same multiple of FD, that AE
is of CF ; therefore AE is (1. 5.) the same Aul-
tiple of CF, that EG is of CD ; but AE, by the
hypothesis, is the same multiple of CF that AB
is of CD, therefore EG is the same multiple of
CD that AB is of CD ; wherefore EG is equal to
AB (1. Ax. 5.). Take from them the common
magnitude AE ; the remainder AG is equal to
the remainder EB. Wherefore, since AE is the
same multiple of CF, that AG is of FD, and that
AG is equal to EB; therefore AE is the same
multiple of CF, that EB is of FD : but AE is the
same multiple of CF, that AB is of CD ; therefore EB is the same
multiple of FD, that AB is of CD. Therefore, if any magnitude, &c.
a. E.D.

E—

PROP. VI. THEOR.

Ii- two magnitudes be equimultiples of two others, and if equi-
multiples of these be taken from the first two, the remainders are
either equal to these others, or equimultiples of them.*

Let the two magnitudes AB, CD be equimultiples of the two E, P,
and AG, CH taken from the first two be equimultiples of the same
E, F ; the remainders GB, HD are either equal to E, F, or equimulti-
ples of them.

•See Note.

• *

BOOK ▼.

THB BLCMBMTfi OF EUCLro.

101

First, let GB be equal to K, HD is equal
to F : make GK equal to F ; and because
AG is the same multiple of E, that CH is
of F, and that GB is equal to E, and CK
to F ; therefore AB is the same multiple
of E, that EH is of F. But AB, by the
hypothesis, is the same multiple of E that
CD is of F; therefore KH is the same
multiple of F, that CD is of F ; wherefore
KH is equal to CD (1. Ax. 5.) : take away
the common magnitude CH, then the re-
mainder KC is equal to the remainder HD ;
but KC is equal to F; HD therefore is
equal to F.

But let GB be a multiple of E : then
HD is the same multiple of F : make CK
the same multiple of F, that GB is of E:
and because AG is the san\e multiple of
E, that GH is of F ; and GB the same
multiple of E that CK is of F ; therefore
AB is the same multiple of E, that KH is
of F (2. 5.) : but AB is the same multiple
of E, that CD is of F, therefore KH is the
same multiple of F, that CD is of it :
wherefore KH is equal to CD (1. Ax. 5.) :
take away CH from both: therefore
the remainder KC is equal to the re-
mainder HD : and because GB is the
same multiple of E, that KC is of F, and
that KC is equal to HD ; therefore HD is
the same multiple of F, that GB is of E.
tudes, &c. d. E. D.

PROP. A. THEOR.

H—

DBF

H—

D E F

If therefore two magni-

If the first of four magnitudes have to the second the same
ratio which the third has to the fourth; then, if the first be
greater than the second, the third is also greater than the fourth;
and if equal, equal ; if less, less.*

Take any equimultiples of each of them, as the doubles of each ;
then, by def 5th of this book, if the double of the first be greater
than the double of the second, the double of the third is greater than
the double of the fourth; but if the first be greater than the second,
the double of the first is greater than the double of the second ;
wherefore also the double of the third is greater than the double of
the fourth ; therefore the third is greater than the fourth : in like
manner, if the first be equal to the second, or less than it, the third
can be proved to be equal to the fourth, or less than it Therefore,
if the first, &c. Q. E. D.

* See Notes.

TAE KLBMBirrS OF SCJQLIO.

BOOK V»

PROP. B. THEOR.

If four magnitudes be proportionals they are popordonals also
when taken inversely.*

If the magnitude A be to B, as C is to D, then also inversely B is
to A, as D, to C.

Take of B and D any equimultiples whatever
£ and F ; and of A and C any equimultiples
whatever G and H. First, let E be greater than
6, then G is less than E; and because A is to
B as C is to D, and of A and C, the first and
third, G and H are equimultiples; and of B and
D, the second and fourth, E and F are equimul-
tiples ; and that G is less than E, H is also (5.
def. 5.) less than F; that is, F is greater than GAB £
H ; if therefore £ be greater than G, F is H C D F
greater than H; in like manner, if E be equal
to G, F may be shown to be equal to H ; and,
if less, less ; and E, F are any equimultiples
whatever of B and D, and G, H any whatever
of A and C ; therefore as B is to A, so is D to
C. If then, four magnitudes, <Stc. cL E. D.

PROP. C. THEOR.

If the first be the same multiple of the second, or the same
part of it, that the third is of the fourth ; the first is to the second,
as the third is to the fourth.*

Let the first A be the same multiple of B the
second that C the third is of the fourth D : A
is to B as C is to D.

Take of A and C any equimultiples whatever
£ and F ; and of B and D any equimultiples
whatever G and H; then, because A is the same
multiple of B that G is of D ; and that E is the
same multiple of A that F is of C ; £ is the
same multiple of B that F is of D (3. 5.) ; there- A
fore E and F are the same multiples of B and E
D : but G and H are equimultiples of B and D :
therefore, if E be a greater multiple of B, than
G is, F is a greater multiple of D, than H is of
D: that is, if E be greater than G, F is greater
than H : in like manner, if £ be equal to G, or
less ; F is equal to H, or less than it. But £,
F are any equimultiples whatever of A, C, and
G, H any equimultiples whatever of B, D. There-
fore A is to B, as C is to D (5. def.)

C D
G F H

*S^e Notes.

BOOK V.

THB ELEMENTS Of EOCLIB;

I6i

Next, Let the first A be the same part of the
second B, that the third C is of the fourth D : A
is to B, as C is to D: for B is the same multi-
ple of A, that D is of G: wherefore, by the pre-
ceding case, B is to A, as D is to C ; and in-
versely (B. 6.) A is to B, as C is to D. There-
fore, if the first be the same multiple, &c. Q.
RD.

A B

PROP. D. THEOR.

If the first be to the second as the third to the fourth, and if
the first be a multiple, or part of the second ; the third is the same
multiple, or the same part of the fourth.*

Let A be to B, as C is to D ; and first let A be a multiple of B, C
is the same multiple of D.

Take E equal to A, and whatever multiple
A or £ is of B, make F the same multiple of
D : then, because A is to B, as C is to D ; and
of B the second, and D the fourth equimultiples
have been taken E and F ; A is to E as C to F
(Cor. 4. 5.) : but A is equal to E, therefore C is
equal to F (A. 5) : and F is the same multiple
of D, that A is of B. Wherefore C is the same
multiple of D, that A is of B.

Next, Let the first A be a part of the second E P

B; C the third is the same part of the fourtbt
D.

Because A is to B, as C is to D ; then in^
versely, B is (B. 5.) to A, as D to C : but A is
a part of B, therefore B is a multiple of A ; and,
by the preceding case, D is the same multiple of
C, that is, C is the same part of D, that A is of
6; therefore, if the first, &c. Q^ E. D.

A B C D

PROP. Vn. THEOR.

Equal magnitudes have the same ratio to the same magni-
tude; and the same has the same ratio to equal magnitudes*

Let A and B be equal magnitudes, and C any other. A and B
have each of them the same ratio to C ; and C has the same tatxo to
each of the magnitudes A and B.

Take of A and B any equimultiples whatever D.and E, and of
C any multiple whatever F; then, because D is the same mtthi-

* See Note.

t See the figure above.

104

TAB ELBMSNTa OF EUCLID.

BOOK V.

pie of A, that £ is of B, aBd that A is equal to
B; D is (i. Ax. 5.) equal to E: therefore, if D
be greater than F, E is greater than F : and
if equal, equal ; if less, less : and D, E are any
equimultiples of A, B, and F is any multiple of
C. Therefore (5. def. 5.) as A is to C, so is B
toC.

Likewise C has the same ratio to A, that it
has to B : for having made the same construc-
tion, D may in like manner be shown equal to
E : therefore, if F be greater than D, it is like-
wise greater than E; and if equal, equal; if
less, less : and F is any multiple whatever of
C, and D, E are any equimultiples whatever of
A, B. Therefore, C is to A, as C is to B (5.
def. 5.). Therefore equal magnitudes, &c. Q.
E.D.

A
B

C F

F-

PROP. VIIL THEOR.

Or unequal magnitudes, the greater has a greater ratio to the
same than the less has ; and the same magnitude has a greater
ratio to the less, than it has to the greater.*

Let AB, BC be unequal magnitudes, of which AB is the greater,
and let D be any magnitude whatever: Fig. L

AB has a greater ratio to D than. BC to
D ; and D has a greater ratio to BC than E
toAB.

If the magnitude which is not the great-
er of the two AC, CB, be not less than D,
take EF, FG, the doubles of AC, CB, as in
Fig. 1. Byt, if that which is not the great-
er of the two AC, CB be less than D (as
in Fig. 2. and 3.) this magnitude can be
multiplied, so as to become greater than
D, whether it be AC, or CB. Let it be G
multiplied, until it become greater than D ;
and let the other be multiplied as oflen ;
and let EF be the multiple thus taken of
AC, and FG the same multiple of CB;
therefore EF and FG are each of them
greater than D : and in every one of the
cases, take H the double D, K, its triple, and so on, till the multiple
of D be that which first becomes greater than FG : let L he that mnl-
tiple of D which is first greater than FG, and K the multiple of D
which is next less than L.

K H D
I

* See Note*

BOOK V.

THS BLIMSNTS OF SVCUB*

ie&

Then, because L is the multiple of D, which is the first that be*
comes greater than FG; the next preceding multiple K is not
greater than FG ; that is, FG is not less than K : and since £F is
the same multiple of AC, that FG is of CB ; FG is the same mul-
tiple of CB, that EG is of AB (1. 5.}: wherefore EG and FG are
equimultiples of AB

Fig. 2.
E

Fig. 3.

F--

G B
L K

and CB: and it was
shown, that FG was
not less than K, and,
by the construction,
EF is greater than D;
therefore the whole A

EG is greater than K ^1

and D together; but, ^

K together with D, is
equal to L; therefore
EG is greater than L ;
but FG is not greater
than L; and EG, FG
are equimultiples of
AB, BC, and L is a
multilple of D ; there- L K H D
fore (7. def. 6.) AB
has to D a greater ra-
tio than BC has to D.
Also D has to BC a
greater ratio than it
has to AB, for, having
made the same con-
struction, it may be
shown, in like manner,

that L is greater than |

FG, but that it is not
greater than EG : and L is a multi^e of D $ and FG, EG are equi'
multiples of CB, AB; therefore D has to CB a greater ratio (7. def.
5.) than it has to AB. Wherefore, of unequal magnitudes, &a Q.
RD.

G B

L K D

PROP. IX. THEOR.

M AGNfTUDEs whlch have the same ratio to the same magnitude
are equal to one another ; and those to which the same magni-
tude has the same ratio are equal to one another.^ ^

Let A, B have each of them the same ratio to C : A is equal to
B : for if they be not equal, one of them is greater than the other j
let A be the greater y then, by what was i^own ia the preceding

* See Note.

105

THE ELSMfiNTB OF &6CUD.

BOOfc V.

proposition), there are some eqtiiihuItlfAes of A and B, and some
multiple of C such, that the multiple of A is greater than the multiple
cf C, but the multiple of B is not gfeater thfitn that of C. Let such
multiples be takeh, and let D, E, be the equimuftiptes of A, B, and
P the multiple of C, so that D may be greater than F*, and E not
greater than F : but, because A is to C, as B is to C, and of A, B
are taken equimultiples D, E, and of C
is taken a multiple F; and that D is
greater than F ; E shall also be greater
than F (5. def. 5.): but E is not greater
than F, which is impossible; A there- A
fore and B are not unequal ; that is,
they are equat.

Next, let G have the same ratio to
each of the magnitudes A and B ; A is
equal to B : for if they be not, one of
them is greater than the other ; let A B
be the greater"; therefore, as was shown
in Prop. 6th, there is some multiple F
of C, and some equimultiples E and D,
of B and A such, that F is greater than E, and not greater than D ;
but because C is to B, as C is to A, and, that F, the multiple of the
first, is greater than E, the multiple of the second ; F the multiple
of the third, is greater than D, the multiple of the fourth (5. def. 6.) :
but F is not greater than D, which is impossible. Therefore A is
equal to B. Wherefore, magnitudes which, &c. Q. K D.

PROP. X. THEOR.

That magnitude which has a greater ratio than another has
unto the same magnitude, is the greater of the two : and that
magnitude, to which the same has a greater ratio than it has
unto another magnitude is the lesser of the two.*

Let A have to C a greater ratio than B has to C : A is greater
than B : for, because A has a greater ratio to C, than B has to O,
there are (7. def. 5.) some equimultiples of A and B, and some
multiple of C such, that the multiple of A is greater than the mul-
tiple of C, but the multiple of B, is not greater than it : let them

* See Note.

1K>OC V^

THA SUHfffiNTS OF EUCLID.

107

be taken, and let D, E be equimulti-
l^s oi A, B, and F a multiple of C
such, that D is greater than F, but E
is not greater than F : therefore D is
greater than E : and, because D and
£ are equimultiples of A and B, and
D is greater than E ; therefore A is
(4. Ax. 5.) greater than B.

Next, let C have a greater ratio to
B that it has to A ; B is less than A :
for there is some multiple F of C, and
some equimultiples E and D of B and
A such, that F is greater than E, but
it is not greater than D : E therefore
is less th^ D; and because E and
D are equimultiples of B and A, there-
fore B is (4. Ax. 5.) less than A.
' That magnitude therefore, &c. CI.
E. D.

PROP. XL THEOR.

Ratios that are the same to the same ratio, are the same to
one another.

Let A be to B as C is to D ; and as C to D, so let E be to F ; A is
to B as E to F.I

Take of A, C, E any equimultiples whatever G, H, K ; and of 6,
D, F any equimultiples whatever L, M, N. Therefore, since A is to
B, as C to D, and G, H are taken equimultiples of A, C, and L, M
of B, D ; if G be greater than L, H is greater than M ; and if equal,
equal; and if less, less (5. def. 6.). Again, because C is to D, as E
to F, and H, K are taken equimultiples of C, E: and M, N, of D, F :
if H be greater than M, K is greater than N ; and if equal, equal ;
and if less, less: but if G^ greater than L, it has been shown that

G-

H-

K-

E-

B-

D-

M-

F-

N-

H is greater than M ; and if equal, equal ; and if less, less: therefore,
if G be greater than L, K is greater than N ; and if equal, equal ;
and if less less : and G, K are any equimultiples whatever of A, E ;
and L, N any whatever of B, F : therefore as A is to B, so is E to F
(5. de£ 5.). Wherefore ratios that, ioc. €t E. D.

108 THB fiLfiMKirra op cuclid. book ▼.

PROP. Xn. THEOR.

If any number of magnitudes be proportionals as one of the
antecedents is to its consequent, so shall all the antecedents taken
together be to all the consequents.

Let any number of magnitudes A, B, C, D, E, F be proportionals ; that
is, as A is to B, so is C to D, and £ to F : as A is to B, so shall A,
C, E together be to B, D, F together.
Take of A, C, E any equimultiples whatever G, H, K ; and of B,

G H K-

A O' 111 I B»

B D F-

M N-

D, F any equimultiples wh^itever, L, M, N : then because A is to B,
as C is to D, and E to F ; and that G, H, K are equimultiples of A,
C, E, and L, M, N equimultiples of B, D, F; if G be greater than L,
H is greater than M, and K greater than N ; and if equal, equal ; and
if less, less (5. def. 5.). Wherefore, if G be greater than L, then G,
H, K together are greater than L, M, N, together ; and if equal, equal ;
and if less, less. And G, and G, H, K, together are any equimulti-
ples of A, and A, G, E together ; because, if there be any number
of magnitudes equimultiples of as many, each of each, whatever
multiple one of them is of its part, the same multiple is the whole
of the whole (1. 5.) : for the same reason L, and L, M, N are any equi-
multiples of B, and B, D, F : as therefore A is to B, so are A, C, E
together to B, D, F together. Wherefore, if any number, &c. CI.

E. D.

PROP. Xra. THEOR.

If the first has to the second the same ratio which the third
has to the fourth, but the third to the fourth a greater ratio than
the fifth has to the sixth ; the first shall also have to the second a
greater ratio than the fifth has to the sixth.*=

Let A the first have the same ratio to B the second, which C the
third, has to D the fourth, but the third to D the fourth, a greater
ratio than E the fifth to F the sixth : also the first A shall have to
the second B a greater ratio than the fifth E to the sixth F.

Because C has a greater ratio to D, than E to F, there are some
equimultiples of C and F, and some of D and F, such, than the
multiple of C is greater than the multiple of D, but the multiple

*SecNote.

BOOic y.
M —

TRS .elsmjbhts of iucud.

109

G-

H-

fi.

E-
F.

N-

K-

of E is not greater than the multii^e of F (7. def. 5.) ; let such be
taken, and of C, £ let G, H be equimultiples, and K, L equimultiples
of D, F, so that G be greater than K, but H not greater than Lr; and
whatever multiple G is of C, take M the same multiple of A ; and
whatever multiple K is of D, take N the same multiple of B; then,
because A is to B, as to D, and of A and C, M and G are equi-
multiples : and of B and D, N and K are equimultiples ; if M be greater
than N, G is greater than K, and if equal, equal ; and if less, less (5.
def 6.) : but G is greater than K, therefore M is greater than N : but
H is not greater than L ; and M, H are equimultiples of A, E ; and
N, L equimultiples of B, F : therefore A has a greater ratio to B, than
E has to F (7. def. 6*). Wherefore, if the first, &c. a. E. D.

Cor. And if the first have a greater ratio to the second, than the
third has to the fourth, but the third the same ratio to the fourth,
which the fifth has to the sixth : it may be demonstrated, in like
manner, that the first has a greater ratio to the second, than^he filth
has to the sixth.

PROP. XIV. THEOR.

If the first has to the second, the same ratio which the third
has to the fourth ; then, if the first be greater than the third, the
second shall be greater than the fourth ; and if equal, equal ; and
if less, less.*

Let the first A have to the second B, the same ratio which the third
C has to the fourth D ; if A be greater than C, B is greater than D.

Because A is greater than C, and B is any other magnitude, A has
to B a greater ratio than C to B (8. 5.); but as A is to B, so is C to

A B C D

ABC

BCD

D ; therefore also C has to D a greater ratio than has to B (13. 5.}:
but of two magnitudes, that to which the same has the greater ratio
is the lesser (10. 5.): wherefore D is less than B ; that is, B is greater
than D.

Secondly, ITA be equal to C, B is equal to D ;. for A is to B, as C,
that is A, to D ; B therefore is equal to D (9. 50.

* See Note.

TH» litBMVMTS OF WOChVB.

BOO* v.

K-

B C

E F

Thirdly, If A be less than C, B shall be less than D ; for C is g^reater
than A, and because C is to D, as A is to B, D is greater than B, by
the first case ; wherefore B is less than D. Therefore, if the first, &c.
Q. £. D.

PROP. XV. THEOR.

Maonitudbs have the same ratio to one another which their
equitnultiples have.

Let AB be the same multiple of C, that DE is of P ; C is to F, as
AB to DE.

Because AB is the same multiple of C, that DE is of P; there are
as many magnitudes in AB equal to C, as ^

there are in DE equal to F : let AB be di-
vided into magnitudes, each equal to C,
viz. AG, GH, HB ; and DE into magni-
tudes, each equal to F, viz. DK, KL, LE :
then the number of the first AG, GH, HB, q--
shall be equal to the number of the last
DK, KL, LE : and because AG, GH, HB
are all equal, and that DK, KL, LE are jj.
also equal to one another ; therefore AG
is to DK, as GH to KL, and as HB to LE
(7. 5.) : and as one of the antecedents to its
consequent, so are all the antecedents to-
gether to all the consequents together (12, 5.) ; wherefore, as AG is
to DK, so is AB to DE : but AG is equal to C, and DK to F : there-
fore, as C is to F, so is AB to DE. Therefore magnitudes, &c. Q. E. D,

PROP. XVL THEOR.

If four magnitudes of the same kind be proportionals, they
shall also be proportionals when taken alternately.

Let the four magnitudes. A, B, C, D, be proportionals, viz. as A to.
B, so C to D: they shall also be proportionals, when taken alternately ;
that is, A is to C, as B to D.

Take of A and B any equimultiples whatever E and F : and of
C and D take any equimultiples whatever G and H : and because
K is the same multiple of A, that F is of B, and that magnitudes
have the same ratio to one another which their equimultiples
have (15. 5.); therefore A isto B, as E Is to F: but as A is to B
so is C to D : wherefore

as C is to D, so (U. 5.) E G

is E to F : again, because

G, H are equimultiples A " > C

of C, D, as C is to D, so
is G toH(15. 5.); but
as C is to D, so is E to
F. Wherefore as E is
toF, so is G to H (IL
5.). But, when four magnitudes are proportionals, if the first be

F-

■*^

H-

ROOK V,'

THB ELSMfiNfB Of EUCUO.

Ill

greater than the tbirdt the second shall be greater than the fourth ;
and if equal, equal ; if less, less (14. 5.)* Wherefore, if E be greater
than 6, F likewise is greater than H ; and if equal, equal ; if less,
less ; and E, F are any equimultiples whatever of A, B ; and G, H
any whatever of C, D. Therefore A is to C, as B to D (5. de£ 5.).
If then four magnitudes, &c. Q. E. D.

PROP. XVn. THEOH.

If magnitudes, taken jointly, be proportionals, they shall also
be proportionals when taken separately ; that is, if two magni*
tudes together have to one of them the same ratio which two
others have to one of these, the remaining one of the first two
shall have to the other the same ratio which the remaining one
of the last two has to the other of these.*

Let AB, BE, CD, DF be the magnitudes taken jointly which are
proportionals ; that is, as AB to BE, so is CD to DF ; they shaU
also be proportionals taken separately, viz. as AE to EB, so CF to
FD.

Take of AE, EB, CF, FD, any equimultiples wh&tever of GH, HK,
LM, MN ; and again, of Efi, FD, take any equimultiples whatever
KX, NP : and because GH is the same multiple of AE, that HK is of
EB, therefore GH is the same multiple (1.5.) of AE, that GK is of
AB ; but GH is the same multiple of AE, that LM is of CF: where-
fore GK is the same multiple of AB, that LM is of CF. Again, be-
cause LM is tlie same multiple of CF, that MN is of FD ; therefore
LM is the same multiple (1. 5.) of CF, that LN is of CD : but LM
was shown to be the same multiple of CF, that GK is of AB ; GK
therefore is the same multiple of AB, that LN is of CD : that is, GK,
LN are equimultiples of AB, CD. Next, because HK is the same
multiple of EB, that MN is of FD ; and
that KX is also the same multiple of EB, X
that NP is of FD ; therefore HX is the
same multiple (2. 5.) of EB, that MP is
of FD. And because AB is to BE, as
CD is to DF, and that of AB and CD,
GK and LN are equimultiples, and of
EB and FD, HX and MP are equimul-
tiples ; if GK be p^eater than HX, then
LN is greater than MP ; and if equal,
equal ; and if less, less (5. def. 5.) : but B D

if GH be greater than KX, by adding
the common part HK to both, GK lis
greater than HX; wherefore also LN
is greater than MP; and by taking
away MN from both, LM is greater G
than NP ; therefore, if GH be greater

K—

H....

N-

M-

E-L

« See Note.

113

THE ELBMSNl^S OF EUCLID.

BOOK V.

than KX, LM is greater than NP. In like manner it may be demonr
strated, that if GH be equal to KX, LN likewise is equal to NP ; and
if less, less : and GH, LM are any equimultiples whatever of AE, CF»
and KX, NP are any whatever of EB, FD. Therefore (6. def. 6.)^ as
AE is to EB, so is CF to FD. If then magnitudes, &c. Q. E. D.

PROP. XVffl. THEOR. .

If magnitudes, taken separately, be proportionals, they shall
also be proportionals when taken jointly, that iSf if the first be to
the second, as the third to the fourth, the first and second to-
gether shall be to the second, as the third and fourth together to
3ie fourth.*

Let AE, EB, CF, FD be proportionals ; that is, as AE to EB, so
is CF to FD : they shall also be proportionals when taken jointly ;
that is, as AB to BE, so CD to DF.

Take of AB, BE, CD, DF any equimultiples whatever GH, HK,
LM, MN ; and again, of BE, DF, take any whatever equimultiples
KO, NP ; and because KO, NP are equimultiples of BE; DF ; aad
that KH, NM are equimultiples likewise of BE, DF, if KO, the mul-
tiple of BE, be greater than KH, which is a multiple of the same BE^
NP, likewise the multiple of DF, shall be greater than NM, the mul-
tiple of the same DF : and if KO be
equal to KH, NP shall be equal to
NM; and if less, less.

First, Let KO not be greater than
KH, therefore NP is not greater than M

NM ; and because GH, HK are equi- „ |

multiples, of AB, BE, and that AB
is greater than BE, therefore GH
is greater (3. Ax. 5.) than KH : but
KO is not greater than KH, where-
fore GH is greater than. KO. In like
manner it may be shown, that LM
is greater than NP. Therefore if
KO be not greater than KH, then
GH, the multiple of AB, is always
greater than KO, the multiple of BE;
and likewise LM, the multiple of CD, G
greater than NP, the multiple of DF.

Next, let KO be greater than KH: therefore, as has been
shown, NP is greater than NM: and because the whole GH is
the same multiple of the whold AB, that HK is of BE> the re-

H
04-

K—

B
E-

* Sec Note.

BOOK V.

THB ELEMENTS OP EUCLID.

113

O
jiii. ..

K—

JV-

E-

mainder OK is the same multiple of
the remainder AE that GH is of AB
(5. 5.) : which is the same that LM
is of CD. In like manner, because
LM is the same multiple of CD, that
MN is of DF, the remainder LN is
the same multiple of the remainder
OF, that the whole LM is of the
whole CD (5. 5.) : but it was shown
that LM is the same multiple of CD,
that GK is of AE ; therefore GK is
the same multiple of AE, that LN is
of CF ; that is, GK, LN are equimul-
tiples of AE, CF : and because KO,
NP are equimultiples of BE, DF, if
from KO, NP there be taken KH,
NM, which are likewise equimultiples of BE, DF, the remainders
HO, MP are either equal to BE, DF, or equimultiples of them (6. 5.).
First, let HO, MP, be equal to BE, DF ; and because AE is to EB,
as CF to FD, and that GK, LN are equimultiples of AE, CF ; GK
shall be to EB, as LN to FD (Cor. 4. 5.) : but HO is equal to EB,
and MH to FD ; wherefore GK is to HO as LN to MP. If therefore
GK be greater than HO, LN is greater than MP ; and if equal, equal;
and if less (Ax. 5.)t less.

But let HO, MP be equimultiples of EB, FD ; and because AE is
to EB, as CF to FD, and that of AE, CF are taken equimultiples GK,
LN, and of EB, FD, the equimultiples HO, MP ; if GK be greater than
HO, LN is greater than MP ; and if
equal, equal; and if less, less (5. def.
5.) : which was likewise shown in the
preceding case. If therefore GH be
greater than KO, taking KH from
both, GK is greater than HO; where-
fore also LN is greater than MP ; and,
consequently, adding NM to both,
LM is greater than NP : therefore, if
GH be greater than KO, LM is great-
er than NP. In like manner it may
be shown, that if GH be equal to KO,
LM is equal to NP ; and if less, less.
And in the case in which KO is not
greater than KH, it has been shown
that GH is always greater than KO, and likewise LM than NP : but
GH, LM are any equimultiples of AB, CD, and KO, NP are any what-
ever of BE, DF : therefore (5. def 5.), as AB is to BE, so is CD to
DF. If then magnitudes, &c. Q. E. D.

O
H—

E—

D
F-

PROP. XIX. THEOR.

If a whole magnitude be to a whole, as a magnitude taken

F-

114 THE CLBMBNTS OF KUGLID. BOOK Y.

from the first, is to a magnitude taken from the other ; the re-
mainder shall be to the remainder, as the whole to the whole.*

IM the whole AB be to the whole CD, as AK, a magnitude taken
from AB, to CF, a magnitude taken from CD ; the remainder BB
shall be to the remainder FD, as the whole AB to the whole CD.

Because AB is to CD, as A£ to CF^ likewise alternately (16. 5.)
BA is to AE, as DC to CP ; and because,.if magni* A
tudes, take jointly, be proportionals, they are also pro- I
portionals (17. 5.) when taken separately ; therefore, ^ |
as BB is to DF, so is EA to FC ; and alternately, as C

BE is to EA, so is DF to FC : but, as AE to CF, so, • ^ |

by the hypothesis, is AB to CD ; therefore also BE, the
remainder shall be to the remainder DF, as the whole
AB to the whole CD : Wherefore, if the whole, &o.
Q. E. D.

Cor. If the whole be to the whole, as a magnitude taken from the
first, is to a magnitude taken from the other; the remainder likewise
lis to the remainder, as the magnitude taken from the first to that
taken from the other : the demonstration is contained in the pre-
ceding,

PROP. E. THEOR.

If four magnitudes be proportionals^ they are also proportionals
by conversion, that is, the first is to its excess above the secondy
as the third to its excess above the fourth.

EL_

C
F-L

Let AB be to BE, as CD to DF ; then BA is to
AE, as DC to CF.

Because AB is to BE, as CD to DF, by division
(17. 5.), AE is to EB, as CF to JFD, and by inver-
sion (B. 5.), BE is to EA, as DF to FC. Wherefore,
by composition (18. 5.), BA is to AE, as DC is to
CF. K therefore, four, &c. Q. E« D. B

PROP. XZ. THEOR.

Ir there be three magnitudes, and other three^ which, taken
two and two^ have the same ratio ; if the first be greater than
the third, the fourth shall be greater than the sixth ; and if equal,
equal ; and if less, less."^

Let A, B, C be three magnitudes, and D, E, F other three,
which, taken two and two, have the same ratio, viz. as A is to B»

* See Note.

BOOK V.

THB BLEMBNT8 OF ROCLID.

115

so is D to B; and as B to C, so is E to F. if A be
greater than C, B shall be greater than F ; and if
equal, equal ; and if less, less.

Because A is gfeater than C, and B is any other
magnitude, and tiiat the greater has to the same
magnitude a greater ratio thioi the less lias to it
(8. 5.) ; therefore A has to B a greater ratio than
C has to B ; but as D is to £, so is A to B ; there-
fore (13. 5.) D has to E a greater ratio than C to A B C
B; and because B is to C, as B to F, by inversion, DBF
C is to B, as F is to E ; and D was shown to have
to E a greater ratio than C to B ; therefore D has
to E a greater ratio than F to E (Cor. 13. 5.); but
the magnitude which has a greater ratio than an-
other to the same magnitude, is the greater of the two (10. 5.) ; D is
therefore greater than F.

Secondly, Let A be equal to C ; D shall be equal to F: because
A and C are equal to one another,
A is to B as C is to B (7. &.) : but A
is to B, as D to E; and C is to B, as
F to E : wherefore D is to E, as F to
E (11. 5.); and therefore D is equal
toF(9. 5.). A B C A B C

Next, let A be less than C;DDEF DBF
shall be less than F : for C is greater
than A, and as was shown in the first
case, C is to B, as F to E, and in like
manner B is to A, as E to D ; there-
fore F is greater than D, by the first
case ; and therefore D is less than F.
Therefore, if there be three, &c. d. B. D.

PROP, XXI. THEOR.

If there be three magnitudes, and other three, 'which have the
same ratio taken two and two, but in a cross order, if the first
magnitude be greater than the third, thb fourth shall be greater
than the sixth ; and if equal, equal ; and if less, less.^

Let A, B; C be three magnitudes, and D,iB, F other lbi^» which
have ihe same ratio, taken two and two, b|it in a cross. at!(ter, m* a0
A is to B, so is E to F, and as B is to C, so is D to
£• If A be greater than C, D shall be greater than
F; and if equal, equal; and if less, less.

Because A is greater than C, and B is any other
magnitude, A has to B a greater ratio (8. 5.) than
C has to B : but as E to F, so is A to B ; therefore
(13. 5.) B has to F a greater ratio than C to B : and
because B is to C, as D to E, by inversion, C i&to A d C

* See Note.

116

THB CUCMEMTS OF BVCLn>.

BOOK V.

B, as £ to D : and E was shown to have to P a D £ F

greater ratio than C to B ; therefore £ has to F a

greater ratio than E to D (Cor. 13. 5.) ; but the

magnitude to which the same has a greater ratio

than it has to another* is the lesser of the two (10«

5.) ; F therefore is less than D, that is, D is greater

than F.

Secondly, let A be equal to C, D shall be equal to F. Because
A and C are equal, A is (7. 5.) to B,
as C is to B: but A is to B, as E to
F ; and C is to B, as E to D ; where-
fore E is to F, as E to D (11. 5.) ;
and therefore D is equal to F (9. 5.).

Next, let A be less than C; D
shall be less than F ; for C is greater ABC A B C
than A, and, as was shown, CistoDEF DEF
B, as E to D ; and in like manner B
is to A, as F to E ; therefore F is
greater than D, by case first ; and
therefore D is less than F. There-
fore, if there be three, &c. Q. £. D.

PROP. XXn. THEOR.

If there be any number of maffnitvdeSf and as many others,
which, taken two and two in order, have the same ratio ; the
first shall have to the last of the first magnitudes the same ratio
which the first of the others has to the last N« B. This is vsw-
cMy cited hy the words " ex sequali,* or " ex GeqteoJ**

First, let there be three magnitudes A, B, C, and as many others
D, E, F, which taken two and two, have the same ratio, that is, such
that A is to B» as D to £ ; and as B is to C, so is E to F ; A shall
be to C, as D to F.

Take of A and D any equimultiples whatever* G and H ; and of
B and £ any equimultiples what-
ever K and L ; and of C and F any
whatever M and N : then, because
A is to B, as D to E, and that G,
H are equimultiples of 'A, D, and
K, L equUnultiples of B, E; as G

is to K, so is (4. 5.) H to L. For A B C D E P
the same reason, K is to M, as L

* fieeNole.

BOOK T.

THfi BLBMENT8 OP EUCLID.

117

to N : and because th^re are three G K M H L N
magnitudes G, K, M, and other
three H, L, N, which, two and two,
have the same ratio : if G be great-
er than M, H is greater than N;
and if equad^ equal ; and if less, less
(20. 5.) ; and G, H are any equimul-
tiples whatever of A, D, and M, N
are any equimultiples whatever of
C, P. Therefore, (6. deC 6.), as A
is to C, so is D to F.

Next, Let there be four magnitudes, A, B, C, D, and other four, Ef
F, G, H, which, two and two, have the same ratio,
viz. as A is to B, so is E to F, and as B to C, so F
to G ; and as C to D, so G to H : A shall be to D,
as £ to H.

Because A, B, C are three magnitudes, and E, F, G other three,
which, taken two and two, have the same ratio ; by the foregoing
case, A is to C, as E to G. But C is to D, as G is to H ; wherefore
again, by the first case, A is to D, as E to H : and so on, whatever
be the number of magnitudes. Therefore, if there be any number,
&c. a. E. D.

A. B. C. D.
E. F. G. H.

PROP. XXm. THEOR.

If there be any number of magnitudes^ and as many others,
which, taken two and two, in a cross order, have the same ratio;
the first shall have to the last of the first magnitudes the same
ratio which the first of the others has to the last. N. B. This is
usually cited by the words '' ex aequali in pr&partione perttirbata f*
oTj^ex sequo perturbatoJ^*

First, Let there be three magnitudes A, B, C, and other three, D,
£, F, which, taken two and two in a cross order, have the same ratio,
that is, such that A is to B, as E to F; andas Bi8toC,soisD toE:
A is to C, as D to F.

Take of A, B, D any equimultiples whatever G, H, K ; and of Cp
E, F any equimultiples whatever L, M, N ; and because G, H are
equimultiples of A, B, and that mag-
nitudes have the same ratio which
their equimultiples have (15. 6.) ; as
A is to B, so is G to H. And for
the saine reason, as E is to F, so is
M to N; but as A is to B, so is E to
F ; as therefore G is to H, so is M to
N(ll. 5.). And because as B is to ABC D E F

•SeeNofap.

118

THi: SLBMBHTa OF BUGLID.

BOOK V«

G H L

K M N

■

C, so Is D to Ey and that H, K are
equimultiples of B, D, and L, M, of
C, E ; as H is to L, so is (4. 5.) K
to M : and it has been shown, that
G is to H, as M to N ; then, because
there are three magnitudes G» H» L,
and other three, K, M, N, which
have the same ratio taken two and
two in a cross order ; if G be great-
er than L, K is greater than N ; and
if equal, equal ; and if less, less (21.
5.) ; and G, K are any equimultiples whatever of A, D ; and L^ N any
whatever of C, F ; as, therefore, A is to C, so is D to F.

Next, Let there be four magnitudes, A, B, C, D, and other four E,
F, G:, H, which ,taken two and two in a cross order,
have the same ratio, viz. A to B, as G to H ; B to C,
as F to G ; and C to J3, as E to F : A is to D, as E
toR

Because A, B, C are three magnitudes, and F, G, H other three,
which, taken two and two in a cross order, have the same ratio ; by
the first case, A is to C, as F to H : but C is to D, as E is to F ; where-
fore again, by the first case, A is to D, as E to H : and so on, what-
ever be the number of magnitudes. Therefore, if there be any num-
ber, &c. Q. E. D.

PROP. XXIV. THBOR.

A. B. C. D.
E. F. G. H.

Ip the first has to the second the same ratio which the third
has to the fourth ; and the fifth to the second, the same ratio
which the sixth has to the fourth; the first and fifth together shall
have to the second, the same which the third and sixth together
have to the fourth.*

Let AB the first have to C the second, the same ratio which DB
the third has to F the fourth; and le^ BG the fifth, have to C the
seccMid, the same ratio which EH the sixth,
has to F the fourth : AG, the first and fifth G
together, shall have to O the second, the H

same ratio which DH, the third and sixth
together, has to F the fourth.

Because BG is to C, as EH to F; by
inversion, C la tg BG, as F to EH: and
because as AB is to C, so is DE to F;
and as C to BG, so F to EH ; ex sequoH
(22. 5.), AB is to BG, as DE to EH:
and because these magnitudes are pro*
portionals, they shall likewise be pro-
portionals when taken jointly (18. 5.): AC

« See Note.

BO<»n

TR< GLfiiraiiTS or BtrCLIO.

110

as, therefore, AG is to GB, so is DH to HE-; but as GB to C, so is
HE to F. Therefore, ex xquali (22. 5.)> as AG is to C, so is DH to
P. Wherefore, if the first, &c. ft. E. D.

Cor. 1. If the same hypothesis be made as in the proposition, the
excess of the first and fifth shall be to the second, as the excess of
tbe third and sixth to the fourth. The demonstration of this is the
same with that of the proposition, if division be used instead of com-
pofiltion.

CoR. 2, The proposition holds true of two ranks of magnitude*,
wbatever be their number, of which each of the first ranlc has to
the second magnitude the same ratio that the corresponding one of
the second rank has to a fourth magnitude ; as is manifest

PROP. XXV. THEOR.

Ip four magnitudes of the same kind are proportionals, the
greatest and least of them together are greater than the other
two together.

Let the four magnitudes AB, CD, E, F, be proportionals ; viz. AB
to CD, as E to F ; and let AB be the greatest of them, and conse-
quently P the least (A. i& 14. 16.). AB, together with F, are greater
than CD, together with E.

Take AG equal to E, and CH equal to F : then, because as AB
is to CD, so is E to F, and that AG is equal to E, and CH equal
to F; AB is to CD, as AG to CH. And ^
because AB the whole is to the whde
CD, as AG is to CH, likewise the re-
mainder GB shall be to the remainder G--
HD, as the whole AB is to the whole
(19. 5.) CD : but AB is greater than CD,
therefore (A. 5.) GB is greater than HD :
and because AG is equal to E, and CH
to F; AG and F together are equal to
CH and E together. If, therefore, to the
unequal magnitudes GB, HD, of which
GB is the greater, there be added equal magnitudes, viz. to GB the
two AG and F, and CH and E to HD ; AB and F together are
greater tiian CD and E. Therefore, if four magnitudes, &c. Q.

D
Hi

C E F

E. D.

PROP. F. THEOR.

Ratios which aire compounded of the same ratios, are the
same with one another.*

Let A be to B, as D to E ; and B to C, as E to F : the ratio which
is compounded of the ratios of A to B, and B to C,

whkh by the definition of compound ratio, is the ratio — a ti n

of A to C, is the same with the ratio D to F, which by !^ ^ p

the same definition is compounded of the ratios of D I L.

to E, and E to F.

«Cfee Note.

A. B. C.
D. E. F.

190 TBS fiUBMurrs of guglid. book v.

Because there are three magnitudes, A, B, C, and three others
D, E, F, which, taken two and two in order, have the same ratio :
ex SRqualh A is to C, as D to F (22. 5.)*

Next, Let A be to B, as £ to F, and B to C, as D to E ; there-
fore, ex xquali in proportione perturbata (23. 5.), A
is to C, as D to F ; that is, the ratio of A to C, which
is compounded of the ratios of A to B, and B to C, is
the same with the ratio D to F, which is compounded
of the ratios of D to E, and E to F : and in like man-
ner the propositions may be demonstrated, whatever be the number
of ratios in either case.

PROP. G. THEOR.

If several ratios be the same with several ratios, each to each :
the ratio which is compounded of ratios which are the same
with the first ratios, each to each ; is the same with the ratio
compounded of ratios which are the same with the other ratios,
each to each.*

Let A be to B, as E to F ; and C to D, as G to H : and let A be
to B, as K to L ; and C to D, as L to M : then the ratio of K to M,
by the definition of compound ratio, is
compounded of the ratios of E to L,
and L to M, which are the same with
the ratios of A to B, and C to D ; and
as*Eto F,' so let N be to O ; and as G
to H, so let O be to P; then the ratio of N to P is compounded of
the ratios of N to O, and O to P, which are the same with the ratios
of E to F, and G to H : and it is to be shown that the ratio of E to
M, is the same with the ratio of N to P, or that E is to M, as N to P.

Because E is to L, as (A to B, that is, as E to F,^ that is, as) N to
O ; and as L to M, so is (C to D, and so is G to H, and so is) O to
P : ex ssquali (22. 5.) E is to M, as N to P. Therefore, if several
ratios, &c. Q. E. D.

A. B. C. D. E. L. M.
E. F. G. H. N. O. P.

PROP. H. THEOR.

If a ratio compounded of several ratios be the same with a
ratio compounded of any other ratios, and if one of the. first ra-
tios, or a ratk) compounded of any of the firsts be the same with
one of the last ratios, or with the ratio compounded of any of the
last ; then the ratio compounded of the remaining ratios of the
first, or the remaining ratio of the firsts if but one remain, is the
same with the ratib compounded of those remaining of the last,
or with the remaining ratio of the last*

* See Note.

BOOK. V. THE CLEMBNT8 OF fifJCLID. 121

Let the first ratios be those of A to B^ B to C, C to D, D to fi,
aiM} E to F : and ]et the other ratios be those of G to H, H to K,
K to L, and L to M : also, let the ratio of A to F, which is com-
pounded of* the first ratios, be the same

A. B. C. D. E. F.
G. H. K. L, M.

with the ratio of G to M, which is com
pounded of the other ratios; and be-
sides, let the ratio of A to D, which is
compounded of the ratios of A to B, B
to C, C to D, be the same with the ratio of G to K, which is com-
pounded of the ratios of G to H, and H to K, then the ratio com-
pounded of the remaining first ratios, to wit, of the ratios of D to
£, and E to F, which compounded ratio is the ratio of D to F, is
the same with the ratio of K to M, which is compounded of the re-
maining ratios of K to L, and L to M of the other ratios.

Because, by the hypothesis, A is to D, as G to K, by inversion
(B. 5.), D is to A, as K to G ; and as A is to F, so is G to M ;
therefore (22. 5.), ex sequalhD is to F, as K to M. If therefore a
ratio which is, &c» d. E. D.

PROP. K. THEOR.

If there be any number of ratios, and any number of other
ratios such, Aat the ratio compounded of ratios ^vhich are the
same with the first ratios, eacn to each, is the same with the
ratio compounded of ratios which are the same, each to eachi
with the last ratios ; and if one of the first ratios, or the ratio
which is compounded of ratios which are the same with several
of the first ratios, each to each, be the same with one of the last
ratios, or with the ratio compounded of ratios which are the
same, each to each, with several of the last ratios: then the ratio
compounded of ratios which are the same with the remaining
ratios of the first, each to each, or the remaining ratio of the
first, if but one remain ; is the same with the ratio compounded
of ratios which are the same with those remaining of the last,
each to each, or with the remaining ratio of the lastf

Let the ratios of A to B, C to D, E to F, be the first ratios ;
and the ratios of G to H, K to L, M to N, O to P, Q, to R, be the
other ratios : and let A be to B, as S to T ; and C to D, as T to
V; and E to F, as V to X: therefore, by the* definition of com-
pound ratio, the ratio of S to X is compounded of the ratios of S

h, k, 1,
A, B; CD; E, F. S, T, V, X.

G, H: K, L; M, N; O, P; ft, R. Y, Z, a, b, c, d.
e, f, g, m, n, o, p.

to T, T to V, and V to X, which are the same with the ratios of A to
B, C to D, E to F, each to each ; also, as G to H, so let Y be to Z ;

* Definition of compound ratio. t See Note.

122 THE ELEMElfTS OP SaCLID. BOOK Y.

and E to L, as Z to a; M to N, as a to b, O to P, as b to c; and d
to R, as c to d: therefore, by the same definition, the ratio of Y to d
is compounded of the ratios of Y to Z, Z to a, a to b, b to c, and c
to d, which are the same, each to each, with the ratios of G to H, K
to L, M to N, O to P, and Q, to R : therefore, by the hypothesis, S is
to X, as Y to d; also, let the ratio of A to B, that is the ratio of
8 to T, which is one of the first ratios, be the same with the ratio
of e to g, which is compounded of the ratios of e to f, and f to g,
which, by the hypothesis, are the same with the ratios of Q to H,
and K to L, two of the other ratios ; and let the ratio of h to 1 be
that which is compounded of the ratios of h to k, and k to 1, whicti
are the same with the remaining first ratios, viz. of C to D, and £ to
F ; also, let the ratio of m to p, be that which is compounded of the
ratios of m to n, n to o, and o to p, which are the same, each to each,
with the remaining other ratios, viz. of M to N, O to P and €1 to R :
then the ratio of h to 1 is the same with the ratio of m to p, or h is
to I, as m to p.

h.k,l.

D: E,F.

S.T,

v.x.

G,H;

N; 0,P;

Q. R. Y, Z,

a, b, c.

e,t

m, n« o.

P-

Because e is to f^ as (G to H, that is, as) Y to Z : and f is to g, as
(K to L, that is, as) Z to a ; therefore, ex atquali, e is to g, as Y to
a : and by the hypothesis, A is to B, that is, S to T, as e to g ; where-
fore S is to T, as Y to a ; and by inversion, T is to S, as a to Y ;
and S Is to X, as Y to d : therefore, ex ssquali, T is to X, as a to d :
also, because h is to k, as (C to D, that is, as) T to V : and k is to 1,
as (E to F, that is, as) V to X ; therefore ex mqtudi, h is to 1, as T
to X : in like manner, it may be demonstrated, that m is to p, as
a to d : and it has been shown, that T is to X, as a to d; tbei^ore
(11. 5.) h is to I, as m to p. Q. E. D.

The propositions G and K are usually, for the sake of brevity, ex-
pressed in the same terms with propositions F and H : and therefore
it was proper to show the true meaning of them when they are so
expressed; especially since they are very frequently made use of by
geometers*

THE

EI^EMENTS OF EUCLID.

BOOK VI.

DEFINITIONS.

Similar rectilineal figures are those
which have their several angles
equa], each to each, and the sides
about the equed angles proportionals.

IL
•♦Reciprocal figures, viz. triangles and parallelogn^ms are such aa
** have their sides about two of their angles proportionals in such
•* manner, that a side of the first figure is to a side of the other,
*" as the remaining side of this other is to the remaining side of the
•* first."*

A straight line is said to be cut in extreme and mean ratio, when
the whole is to the greater segment, as the greater segment is to
the less.

IV.

The altitude of any figure is the straight line
drawn firbm ks vertex perp^dicular to the
base.

PROP. I. THEOR.

Tbiavgles and parallelograms of the same altitude are to one
another as their bases.*

Let the triangles ABC, ACD, and the parallelograms EC, CF^have
the same altitude, viz. the perpendicular drawn firom the point A ta
BD : then, as the base BC is to the base CD, so is the triangle ABC
to the triangle ACD, and the perallelpgram EC to the paralldQgram
CF.

« See Note.

124 THE ELEMENTS OP EUCLID. BOOK VI-

Produce BD both ways to the points H, L, and take any number
. of straight lines BG, GH, each equal to the base EC ; and DK, KL,
any number of them, each equal to the base CD, and join AG, AH,
AK, AL : then, because CB, BG, GH are all equal, the triangles AHG,
AGB, ABC are all equal (38. 1.): therefore, whatever multiple the
base HC is of the base BC, the same multiple is the triangie AHC of
the triangle ABC ; for the same reason, whatever multiple the base
LC is of the base CD, the E A F

same multiple is the triangle
ALC of the triangle ADC : and
if the base HC be equal to the
base CL, the triangle AHC is
also equal to the triangle ALC
(38. 1.) ; and if the base HC be
greater than the base CL, like-
wise the triangle AHC is great- H G B C D K L
er than the triangle ALC ; and if less, less: therefore, since there are
four magnitudes, viz. the two bases BC, CD, and the two triangles
ABC, ACD, and of the base BC and the triangle ABC the first and
third, any equimultiples whatever have been taken, viz. the base HC
and the triangle AHC ; and of the base CD and triangle ACD, the
second and fourth, have been taken any equimultiples whatever, viz.
the base CL, and triangle ALC ; and that it has boen shown, that, if
the base HC be greater than the base CL, the triangle AHC is great-
er than the triangle ALC ; and if equal, equal ; and if less, less ; there-
fore (5. def. 5.) as the base BC is to the base CD, so is the triangle
ABC to the triangle ACD.

And because the parallel(^ram CE is double of the triangle ABC
(4L L) and the parallelogram CF double of the triangle ACD, and
that magnitudes have the same ratio which their equimultiples have
(15. 5.) ; as the triangle ABC is to the triangle ACD, so is the paral-
lelogram EC to the parallelogram CF; and because it has been shown,
that as the base BC is to the base CD, so is the triangle ABC to the
triangle ACD ; and as the triangle ABC to the triangle ACD, so is
the parallelogram EC to the parallelogram CF ; therefore as the base
BC is to the base CD, so is (11. 5.) the parallelogram EC to the
parallelogram CF. Wherefore, triangles, &c. Q. E. D.

Cor. From this it is plain, that triangles and parallelograms that
have equal altitudes, are one to another as their bases.

Let the figures be placed so as to have their bases in the same
straight line ; and having drawn perpendiculars from the vertices of
the triangles to the bases, the straight line which joins the vertices is
parallel to that in which their bases are (33. L), because the perpen-
diculars are both equal and parallel to one another: then, if the same
construction be made as in the proposition, the demonstration will be
the same.

BOOK VI.

THE fiLEMBMTS OF KUCLID.

125

PROP. n. THEOB.

If a straight line be drawn parallel to one of the sides of a
triangle, it snail cut the other sides, or those produced, propor-
tionally ; and if the sides, or the sides produced, be cut propor-
tionally, the straight line which joins the points of section snail
be parallel to the remaining side of the triangle.*

Let DE be drawn parallel to BC, one of the sides of the 'triangle
ABC ; BD is to DA, as CE to EA.

Join BE, CD ; then the triangle BDE is equal to the triangle
CDE (37. 1.), because they are on the same base DE, and between
tbe same parallels DE, BC; ADE is another triangle, and equal
magnitudes have to the same the same ratio (7. 5.) ; therefore, as
the triangle BDE to the triangle ADE, so is the triangle CDE to the
triangle ADE; but as the triangle BDE to the triangle ADE, so i»
(I. 6.) BD to DA, because having the same altitude viz. the perpen-
dicular drawn from the point E to AB, they are to one another as
their bases ; and for the same reason, as the triangle CDE to the
triangle ADE, so is CE to EA. Therefore, as BD to DA, so is CE

toEA(ll. 6.).
Next let the sides AB, AC of the triangle ABC, or these produced,^

be cut proportionally in the points D, E, that is, so that BD be to DA
as CE to EA, and join DE; DE is parallel to BC.

The same construction being made ; because as BD to DA, so ts
CE to EA ; and as BD to DA, so is the triangle BDE to the triangle
ADE (1. 6.) ; and as CE to EA, so is the triangle CDE to the trian-
gle ADE ; therefore the triangle BDE ia to the triangle ADE, a» the
triangle CDE to the triangle ADE ; that is, the triangles BDE, CDE
have the same ratio to the triangle ADE ; and therefore (9. 6.) the
triangle BDE is equal to the triangle CDE ; and they are on the
same base DE ; but equal triangles on the same base are between
the same parallels (39. L) ; therefore DE is parallel to BC, Where-
fore, if a straight line, &c. Q,. E. D.

• See Note.

126 THE BLiai£NrS OF EUCLID. BOOK VI.

PROP. m. THEOR.

If the angle of a triangle be divided into two equal ansles, by
a straight line which also cuts the base; the segments of the base
shall have the same ratio which the other sides of the triangle
have to one another : and if the segments of the base have the
same ratio which the other sides of the triangle have to one
another, the straight line drawn from the vertex to the point of
section, divides the vertical angle into two equal angles.*

Let the angle BAG of any triangle ABC be divided into two equal
angles by the straight line AD : BD is to DC, as BA to AC.

Through the point G draw GE parallel (31. 1.) to DA, and let BA
produced meet CE in E. Because the straight line AG meets the
parallels AD, EC, the angle ACE is equal to the alternate angle
CAD (29. 1.) ; but GAD, by the hypothesis is equal to the angle
BAD; wherefore BAD is equal to the angle ACE. Again, be-
cause the straight line BAE meets the parallels AD, EC, the
outward angle BAD is equal ^ ^ ®

to the inward and opposite a^ ^-^ I

angle AEG : but the angle ACE .^r I

has been proved equal to the an-
gle BAD ; therefore also AGE is
equal to the angle AEG, and con-
sequentl3r the side AE is equal
the side ((5. 1.) AC ; and because g
AD is drawn parallel to one of
the sides of the triangle BCE, viz. to EG, BD is to DC, as BA to
AE (2. 6.) : but AE is equal to AC ; therefore, as BD to DC, so is
BA to AC (7. 5.),

Let now BD be to DC, as BA to AC, and join AD ; the angle BAG
is divided into two equal angles by the straight line AD.

The same construction being made ; because, as BD to DC, so is
BA to AC : and as BD to DC, so is BA to AE (2. 6.), because AD is
parallel to EC ; therefore BA is to AC, as BA to AE, (11.5.): con-
sequently AC is equal to AE (9. &), and the angle AEC is therefore
equal to the angle ACE (5. L) : but the angle AEC is equal to the
outward and opposite angle* BAD: and the angle AGE is equal to
the alternate angle CAD (29. 1.) : wherefore also the angle BAD is
equal to the angle CAD : therefore the angle BAG is cut into two
equal angles by the straight line AD. Therefore, if the angle, &c.
Q.E.D.

* See Note.

BOOK VI. THB BLEMBNTS OF COCLID. 127

PROP. A. THEOR.

If the outward angle of a triangle made by producing one of
its sides, be divided ihto two equal angles, by a straight line
which also cuts the base produced ; the segments between the
dividing line and the extremities of the base have the same
ratio which the other sides of the triangle have to one another:
and if the segments of the base produced have the same ratio
which the other sides of the triangle have, the straight line drawn
from the vertex to the point of section divides the outward angle
of the triangle into two equal angles.

Let the outward angle CAE of any triangle ABC be divided into
two equal angles by the straight line AD which meets the base pro-
duced in D : BD is to DC, as BA to AC.

Through C draw CF parallel to AD (31. 1.): and because the
straight line AC meets the parallels AD, FC, the angle ACF is equal
to the alternate angle CAD (29. 1.) : but CAD is equal to the angle
DAE (Hyp.) ; therefore also DAE is equal to the angle ACF. Again,
because the straight line FAE meets the parallels AD, FC, the out-
ward angle DAE is equal to the E
inward and opposite angle CFA :
but the angle ACF has been proved
equal to the angle DAE; there-
fore also the angle ACF is equal
to the angle CFA, and conse-
quently the side AF is equal to
the side AC (6. 1.): and because ^
AD is parallel to FC, a side of the
triangle BCF, BD is to DC, as BA to AF (2. 6.) : but AF is equal to
AC ; as therefore BD is to DC, so is BA to AC.

Let now BD be to DC, as BA to AC, and join AD ; the angle CAD
is equal to the angle DAE.

The same construction being made, because BD is to DC, as BA
to AC ; and that BD is also to DC, as BA to AF (U. 5.) ; therefore
BA is to AC, as BA to AF (9. 5.) ; wherefore AC is equal to AF (5.
1.), and the angle AFC equal (5* L) to the angle ACF; but the angle
AFC is equal to the outward angle EAD, and the angle ACF to the
alternate angle CAD : therefore also EAD is equal to the angle CAD.
Wherefore, if the outward, &c. Q,. E. D.

PROP, IV. THEOR.

The sides about the equal angles of equiangular triangles are
proportionals; and those which are opposite to the equal angles

128 THE ELEMENTS OF EUCLID. BOOK VI.

are homologous sides, that is, are the antecedents or consequents
of the ratios.

Let ABC, DCE be equiangular triangles, having the angle ABC
equal to the angle DCE, and the angle ACB to the angle DEC, and
consequently (32. 1.) the angle BAC equal to the angle CDE. The
sides about the equal angles of the triangles ABC, DCE are propor-
tionals : and those are the homologous sides which are opposite to
the equal angles.

Let the triangle DCE be placed so that its side CE may be con-
tiguous to BC, and in the same straight line with it : and because
the angles ABC, ACB are together less than two right angles (17.
1.), ABC, and DEC, which is equal F
to ACB, are also less than two right
angles ; wherefore BA, ED produced
shall meet (12. Ax. I.) ; let them be A
produced and meet in the point F ;
and because the angle ABC is equal
to the angle DCS, BF is parallel (28.
1.) to CD. Again, because the angle
ACB is equal to the angle DEC, AC B C £

is parallel to FE (28. 1.) : therefore FACD is a parallelogram ; and
consequently AF is equal to CD, and AC to FD (34. 1.) : and be-
cause AC is parallel to FE, one of the sides of the triangle FBE, BA
is to AF, as BC to CE (2. 6.) : but AF is equal to CD ; therefore
(7. 5.), as BA to CD, so is BC to CE ; and alternately, as AB to BC
so is DC to CE, (17. 1.) : again, because CD is parallel to BF, as
BC to CE, so is FD to DE (2. 6.) : but FD is equal to AC : therefore,
as BC to CE, so i^s AC to DE : and alternately, as BC to CA, ^
CE to DE : therefore, because it has been proved that AB is to BC,
as DC to CE, and as BC to CA, so CE to ED, ex mquali, (22. 5.)
BA is to AC, as CD to DE. Therefore the sides, &c. Q. E. D.

PROP. V. THEOR.

If the sides of two triangles, about each of their angles^ be
proportionals, the triangles shall be equiangular, and have their
equal angles opposite to the homologous sides.

Let the triangles ABC, DEF have their sides proportionals, so that
AB is to BC, as DE to EF ; and BC to CA, as EF to FD ; and con-
sequently, ex ssgiuUif BA to AC, as ED to DF ; the triangle ABC is
equiangular to the triangle DEF, and their equal angles are oppo-
site to the homologous sides, viz. the angle ABC equal to the angle
DEF, and BCA to EFD, and also BAC to EDF.

At the points E, F, in the straight line EF, make (23. 1.) the
angle FEG equal to the angle ABC, and the angle EFGr equal to

BOOK Vl«

THB SLBMSNT8 OF £00.19.

129

BCA; wherefore the remaining
angle BAG is equal to the re-
maining angle EGF (32. !.)« and
the triangle ABC is therefore
equiangular to the triangle
GEF; and consequently they
have their sides opposite to the
equal angles proportionals (4.
6.> Wherefore, as AB to BC, so is GE to EF ; but as AB to BC, so
is DE to EF ; therefore as DE to EF, so (1 1. 5.) GE to EF : there-
fore DE and GE have the same ratio to EF, and consequently are
equal (9. 5.): for the same reason, DF is equal to FG: and because
in the triangles DEF, GEF, DE is equal to EG, and EF common, the
two sides DE, EF are e(iual to the two GE, EF, and the base DF is
equal to the base GF : therefore the angle DEF is equal (8. 1.) to the
angle GEF, and the other angles to the other angles which are sub-
tended by the equal sides (4. 1.) : wherefore the angle DFE is equal
to the angle GFE, and EDF to EGF: and because the angle DEF is
equal to tiie angle GEF, and GEF to the angle ABC ; therefore the
angle ABC is equal to the angle DEF : for the same reason the angle
ACB is equal to the angle DFE, and the angle at A to the angle
at D. Therefore the triangle ABC is equiangular to the triangle DEF,
Wherefore, if the sides, &c. Q. E. D.

PROP. VI. THEOR.

If two triangles have one angle of the one equal to one angle
of the other, and the sides about the equal angles proportionals,
the triangles shall be equiangular, and shall have those angles
equal which are opposite to the homologous sides.

Let the triangles ABC, DEF have the angle BAC in the one equal
to the angle EDF in the other, and the sides about those angles pro-
portionals ; that is, BA to AC, as ED to DF ; the triangles ABC, DEF
are equiangular, and have the angle ABC equal to the angle DEF,
and ACB to DFE.

At the points D, F, in the straight line DF, make (23. 1.) the angle
FDG equal to either of the angles BAC, EDF ; and the angle DFQ
equalto the angle ACB; wherefore A
the remaining angle at B is equal
to the remaining one at G (32. l.)»
and consequently the triangle ABC
is equiangular to the triangle DGF ;
and therefore as BA to AC, so is (4.
ft) GD to DF : but. by the hypothe-

sis, as BA to AC, so is ED to DF •,

as therefore ED to DF, so is (1 1 . 5.) B C E F

GD to DP ; wherefore ED is equal (9. 5.) to DG ; and DP is common

If T»B lLfiMl!lT» OP fittCtlD. tfdolt Vt.

to the two triangles ^DF, GDF ; theralbre the t^o dtdes ED, D9 are
'equal to the two sides GD, DF: and the angle EDF is equal to the
angle GDF ; wherefore the base EF is equal to the base FX> (4. l.}«
and the triangle EDF to the triangle GDF, and the ranaining angles
to the remaining angles, each to each, which are sttbtended by the
equal sides ; there^e the angle DFG is eqnal to the angle DFE, and
the angle at G to the angle at E : but the angle DFG is equal to the
angle ACB ; therefore the angle ACB is equal to ttm angle DFE : and
the angle BAG is equal to the ahgle EDF (Hyp.) ; wherefore also the
retnaiaing angle at B is equal to the remaining angle at E. There^
fore the triangle ABC is equiangular to the triangle DKF* Where-
ibre, if two triangles, &c. 4t E. D.

PROP. VIL THEOa

If two triangles have okie angle of the one equal to one angi^
of the other, and the sides about two other angles proportiontiisv
then, if eacii of the remaining angles be either less^ or not ies&,
than a right angle ; or if one of them be a right atigte ; the trian*
gles shall be equiangular, and have those angles equal about
which the sides are proportionals,*

I^et the two triangles ABC, DEF have one angle in the one equal
to one angle in tlie other, viz. the angle BAC to the angle EDF, and
the sides about two other angles ABC, DEF proportionals, so that
AB is to BC, as DE to EF ; and in the first case, let each of the re<
mamlng angles at C, F be less than a right angle. The triangle ABC
is equiangular to the triangle DEFi viz. the angle ABC is equti to
the angle DEF,, and the remaining angle at C, to the remaining an-*
gle at F.

For, if the angles ABC, DEF be not equal, one of them is greater
than the other: let ABC be the greater, and at the point B, in the
straight line AB, make the angle A

ABG equal to the afigie (23v 1.) ^f p

DEF: end because the angle at ^y^ I

A is equal to the angle at D, and ^^ ] q

the angle ABG to the angle DEF ; ^^^^ / ^

the remaining angle AGB is equal B C E If

(32, 1.) to the remaining angle DFB: therefore the triangle ABG
Is equiangular to the triangle DEF; wherefore, (4. 6.) as AB Is to
BG, so is DE to EF; but as DE to EF, so, by hypothesis, is AB to
BC ; therefore as AB to BC, so is AB to BG (II. 5.) ; and because
AB has the same ratio to each of the lines BC, BG ; BC is eqaal (»•
5.) to BG, and therefore the angle BGC is equal to the angle BCQ
(6. 1.) ; but the angle BCG is, by hypothesis, less than a right an^
gle ; therefore also the angle BGC is less than a right angle, and

» See Note.

900K VI.

THS SLfiMfiNTI OP BUCLIP.

131

the adjacent apgle AGB miiat be greater thin a right angle (19. !.)•
But it was proved that the angle AGB is equal to the angle at F;
therefcNre the angle at F i9 greater than a right angle: but by the
hypothesis, it is less than a right angle which is absurd. Therefore
the angles ABC> DJSF are not uneqiwl, that is, they are equal : and
the angle at A is equal to the angle at D ; wherefore the remainmg
angle at C is equal to the remaining angle at F : therefore the trian-
gle ABC is equiangular to the triangle DEF.

Nezt» let each of the angles at C» F. be not less than a right an*
gle : the triangle ABO is also in this case equiangular to the triangle
DBF.

The same construction being A

made, it may be proved in like yi D

manner that BC is equal to BG| y^V

and the angle at C equal to the y^ \ q

angle BGC: but the angle at C y^,---^

is not less than a right angle; j^ Kr^^T!"^ \
therefore the angle BGC is not leas B C E P

than a right angle ; wherefore two angles of the triangle BGrC are

together not less than two right angles, which is impossible (17. 1.)^

and therefore the triangle ABC may be proved tp be equiangular td

the triangle D£iF, as in the first oase.

Lastly, let one of the angles at C, F, viz. the angle at C, be a right

angle ; in this case likewise the triangle ABC is equiangular to the

triangle DEF.

For, if they be not equiangu- A

lar, make, at the point B of the

straight Ime AB, the angle ABG

equal to the angle DEF; then

it may be proved, as in the first

case, that BG is equal to BC ;

but the angle BCG is a right an-
gle, therefore (5. 1.) the angle

BGC is also a right angle;

whence two of the angles of the

triangle BGC are together not

less than two right angles, which

is impossible (17. 1.); therefore

the triangle ABC is equiangular

to the triangle DEF. Wherefore,

if two triangles, &c. Q,. E. D.

PROP. VBS. 'niBOR.

In a right aisled tripngfe, if a perpendicular be drawn from
the right angle to the base, the tnai^les on each side oi it 9U^
siniilar to the whole triangle, and to one another*^

g** Sm Note.

132 THE £Lratf£NTft OF EUCLID. .AOOK VI.

Let ABC be a right angled triangle, having the right angle BAG ;
and trom the point A let AD be drawn perpendicular to the base
BC : the triangles ABD, ADC are similar to the whde triangle ABC,
and to one another.

Because the angle BAC is equal to the angle ADB, each of them
being a right angle, and that the angle at B is common to the two
triangles ABC, ABD ; the remaining A

angle ACB is equal to the remaining
angle BAD (32. 1.): therefore the
triai^le ABC is equiangular to the
triangle ABD, and the sides about
their equal angles are proportionals
(4. 6.) ; wherefore the triangles are g
similar (1. Def. 6.) ; in the like mEmner
it may be demonstrated, that the triangle ADC is equiangular and
similar to the triangle ABC : and the triangles ABD, ADC, bem^
both equiangular and similar to ABC, are equiangular and similar to
each other. Therefore, in a right angled, &c. d. E. D.

Cor. From this it is manifest, that the perpendicular drawn from
the right angle of a right angled triangle to the base, is a mean
proportional between the segments of the base : and also {hat each
of the sides is a mean proportional between the base, and its seg-
ment adjacent to that side : because in the triangles BDA, ADC, BD
is to DA as DA to DC (4. 6) ; and in the triangles ABC, DBA, BC
is to BA, as BA to BD (4. 6.) ; and in the triangles ABC, ACD, BO
is to CA as CA t6 CD (4. 6.).

PROP. IX. PROB.

From a grven straight line to cut off any part required.*

Let AB be the given straight line ; it is required to cut off any
part from it.

. From the point A draw a straight line AC making any angle with
AB ; and in AC take any point D, and take AC the same multiple
of AD, that AB is of the part which is to be cut off A
Irom it : join BC, and draw DE parallel to it : then
AE is the part required to be cut off.

Because ED is parallel to one of the sides of the e
triangle ABC, viz. to BC, as CD is to DA, so is
(2. 6.) BE to EA ; and, by composition (18. 5.) CA
is to AD as BA to AE : but CA is a multiple of
AD ; therefore (D. 5.) BA is the same multiple of
AE: whatever part therefore AD is of AC, AE is
the same part of AB : wherefore, from the straight ^
line AB the part required is cut off. Which was. to *^
be done.

« See Note.

SOOK ?I.

THS ELSMfiMTS OP EUCLID.

13d

PROP. X, PROB.

To divide a given straight line similarly to a given divided
straight line, that is, into parts that shall have the same ratios to
one another which the parts of the divided given straight line
have.

. Let AB be the straight line given to be divided, and AC the
divided line; it is required to divide AB similarly to AC.

Let AC be divided in the points D, E ; and let AB, AC be placed
so as to contain any angle, and join BC, and through the points
D, E draw (31. 1.) DF EG parallels to it; and through D draw
DHK parallel to AB: therefore each of the figures FH, HB is a
parallelogram; wherefore DH is equal (34. 1.) to FG, and HK
to GB : and because HE is parallel to KC, ' A
one of the sides of the triangle DKC, as CE
to ED, so is (2. 6.) KH to HD: but KH is
equal to BG, and HD to GF ; therefore as
CE to ED, so is BG to GF ; again, because
FD is parallel to EG, one of the sides of the
triangle AGE, as ED to DA, so is GF to
FA ; but it has been proved that CE is to '
ED as BG to GF ; and as ED to DA, so GF ' g
to FA : therefore the given straight line AB
is divided similarly to AC. Which was to be done.

PROP. XI. PROB.

To find a third proportional to two given straight lines.

Let AB, AC be the two given straight lines, and let them be placed
so as to contain any angle ; it is required to find a third proportional
to AB, AC.

Produce AB, AC to the points D, E : and make A
BD equal to AC ; and having Joined BC, through D
draw DE parallel to it (3L 1.).

Because BC is parallel to DE, a side of the tri-
angle ADE, AB is (2. 6.) to BD, as AC to CE : but B
BD is equal to AC ; as therefore AB to AC, so is I
AC to CE. Wherefore, to the two given straight I
lines AB, AC a third proportional CE is found. I
Which was to be dime. jy

PROP. XII. PROB.

To find a fourth proportional to three given straight lines.

Let A, B, C be the three given straight lines ; it is required to find ^
fourth proportional to A, B, C.

..»■

194 THE BLBHBRta OT SOQUD. TOOK Vt.

Take two straight linea DB, DF, ccaitaiBlBg any angle EDF : and

upon these make DG equal to A, GE D

equal to B, and DH equal to C ; and . A

having joined GH, draw BF parallel / \

(ai. 10 to it through the point E: / \ *-

and because GH is parallel to EF, / \ 7
one of the sides of the triangle DEF,

•m-^^^l^^m

II 1 i t

DGlstoGE, asDH to HF(». 6.); p / \ ,,

but DG is equal to A, GE to S, and ^A-*~ — ~^ ' ^
DH to C ; therefore, as A is to B, so
is C to HF : wher^ore to the three

given straight lines A, B, C, a fi>urth B ^

proportional HF is found. Which was to be done.

PROP. XIII. PROB.

To find a mean proportional between two given straight lines-
Let AB, BC be the two given straight lines ; it is required to find a
mean proportional between them.

Place AB, BC in a straight line, and upon AO describe the semi-
circle ADC, and from the point B draw ^^^u ^^J^
(11. 1.) BD at right angles to AC, and y'^^ .XK\
join AD, DC. / ^^ W
Because the angle ADC in a semi- / ^/^ \ \
circle is a right angle (31. 3.), and be- / y^ \ \
cause in the right angled triangle ADC, ly^ \j
DB is drawn from the right angle per- a B ^C
pendicular to the base, DB is a mean

proportional between AB, BC, the segments of the base (Cor. 8. 6.) :
tl^refore between the two given straight lines AB, BC, a mean pro-
portional DB is found. Which was to be done.

PROP. XrV. THEOR.

Equal parallelogranns which have one angle of the one equ^il
to one ansle of thQ other, hav^ their side^ about the equal angles
reciprocally proportional ; and parallelograms that have one an-
gle of the one equal to one angle of the otfaer^ and their sides
a^bout the equal angles reciprocally proportional, are equal to one
another.

Let AB, BC be equal parallelograms, which have the angles at B
equal, and let the i#des DB, BE be placed in the same straight line :
wherefore also FB, BG are in one straight line (14. 1.) : the sides of
the parallelograms AB, BC, about the ^qal ai^lo?) are r^iprocally
proportional ; that is» DB is^ to BE, as GB to BF.

Boox Tt. T8S Humurra or boclid. 135

Compiete the psralldagMni FE : and because the parallelogram
AB is equal to B0« and that FE is an* A F

other paralteiograin, AB Is to FE, as \
BC to FE (7. IS.)t but as AB to FE, so \
is the base DB to Bfi<l. 6.); and «• V.
BC to FE^ so is the base GB to BF :
therefore as DB to BE, so is QB to'BF
(11. 5.). Wherefore the sides of the
paraUdograms AB, BC about their
equal angles are reciprocally propor-
tional.

But, let the sides about the equal .angles be reciprocally propor-
tional, viz. as DB to BE, so GB to BF ; the parallelogram AB is
equal to the parallelogram BC.

Because as DB to BE, so is GB to BF ; and as DB to IBE^ so is
the paraUdogram AB to the parallelogram FE ; and as GB to BF,
so is the paitillelogram BC to the parallelogram FE ; therefore as
AB to FE, so BC to FE (d. 5.) : wherefore the parallelogram AB is
equal (0. 5.) to the parallelogram BC. Therefore, equal parallelo-
grams, &c. a. E. D.

PROP. XV. THEOR.

Eqoal trianffles wfaich have one angle of the one equal to one
angle o( the otner^ have their sides about the equal angte reci*
I»waUy profX)rtional; and triangles which have one angie in Ibt
one equal to one angle in the other, and their sides about the equal
anglei^ reciprocally proportional, are equal to one another.

Let ABC, ADE be equal triangles, which have the angle BAC
equal to the angle DAE ; the sides about the equal angles of the
triangles are reciprocally proportional ; that is, CA is to AD^ as EA
to AB.

Let the triangles be placed so thai thdr sides CA, AD be in one
straight line ; wherefore also EA and AB are in one straight Ixat
(14. 1.) and join BD. Because the triangle ABC is equal to the trl^
angle ADE, and that ABD is another B D

triangle, therefore as the triangle CAB
is to the triangle BAD, so is the trian-
gle EAD to trtengle DAB (7. 5.) : but
«i^ iHtfi^ CAB to ^rfangle BAD, so
is the base CA to AD (1. 6.) : and as
tvlangle EAD to triangle DAB, so is
Hie baise EA to Afi (L a) : as th€ye-
fyr^ CA to AD, 90 is EA to AB {11. O fi

5.) ; wherefom the sides of Ihe triangles ABC, ADE aboitt the ecfas^
an^es are w^tmfcdSiy propc»tSonal.

^ift M the sides of tiie triangles ABC, ADE about the equal angles
be reciproeaHy pi^itioiial, viz. CA to AD, as fiA to AB ; tiie tri-
angle ABC is equal to the triangle ADE.

136

TBI! JItBMBirTa OP BUCLIO.

.8Q0K TI.

Havifig joined BD as before ; because as CA to AD so is EA to
AB ; and as CA to AD, so is triangle BAG to trlangle'lBAD (1. 6.) ;
and as £A to AB, so is triangle EAD to triangle BAD (L 6.) ; there-
fore (11. 5.) as triangle BAG to triangle BAD, so is triangle EAD to
triangle BAD ; that is, the triangles BAG, EAD have the same ratio
to the triangle BAD ; wherefore the triangle ABG is equal (9. 5.)
to' the triangle ADE. Therefore, equal triangles, &c. d. E. D.

PROP. XVI. THEOR.

If four straight lines be proportionals, the rectangle contained
by the extremes is equal to the rectangle contained by the means :
and if the rectangle contained by the extremes be equal to the
rectangle contained by the means/ the four straight lines are
proportionals.

Let the four straight Imes AB, GD, E, F, be proportionals, viz. as
AB to CD, so is E to F ; the rectangle contained by AB, P^ is equal
to the rectangle contained by CD, E. ^

From the points A, C draw (11.1.) AG, CH at right angles to
AB, CD ; and make AG equal to F, and CH equal to E, and com-
plete the parallelograms BG, DH; because as AB to CD, so is E to
F ; and that E is equal to CH, and F to AG ; AB is (7. 5.) to CD,
as CH to AG : therefore the sides of the parallelograms BG, DH
about the equal angles are reciprocally proportional ; but parcdlelo*
grams which have their sides about equal angles reciprocally pro-
portional, are equal to one another (14. 6.) ; therefore the parallelo-
gram BG is equal to the parallelogram DH, and the parallelogram

BG is contained by the straight E —

lines AB, F, because AG is equal H

to F ; and the parallelogram DH is ^
contained by CD and E, because
CH is equal to E; therefore the <
rectangle contained by the straight
lines AB, F is equal to that which
is contained by CD and E.

And if the rectangle contained by
the straight lines AB, F. be equal to a. • B C D

that which is contained by CD, E;
these four lines are proportionals, viz. AB is to CD, as E to F.

The same construction being made, because the rectangle con-
tained by the straight lines AB, F is equal to that which is contained
by CD, E, and that the rectangle BG is contained by AB, F, because
AG is equal to F ; and the rectangle DH by CD, E, because CH is
equal to E; therdbre the parallelogram BG is equal to the parallelo-
gram DH, and they are equiangular : but the sides about the equal
angles of equal parallelograms are reciprocally proportions^ (U. 6.) ;
wber^ore, as AB to CD, so is CH to AG ; and CH is equal to E,
asnd AG to F : as therefore AB is to CD, so E to F. Wherefore, if
four, ^c, d. E. D.

BOOK Yt; na BLsitBrrs of ki^ouik 137

PROP. XVn. THEOR.

If three straight lines be proportionals^ the rectangle contained
by the extremes is equal to the square of the mean : and if the
rectangle contained by the. extremes be equal to the square of the
mean, the three straight lines are proportionals.

Let the three straight lines A, B, G be jmroportionals, viz. as A to
B, 90 B to O ; the rectangle contained by A, C is equal to the square
of B.

Take D equal to B ; and because as A to B, so B to C, and that
B is equal to D ; A is (7. 5.) to B, as D to C ; but if four straight
lines be proportionals, the A -^— — ^— —
rectangle contained by the B — _
extremes is equal to that D '
which is contained by the G

means(16.6.):thereforethe D

rectangle contained by A,
G is equal td that contain*
ed by B, D. But the rect-
angle contained by B, D is . -^ B
the square of B ; because B is equal to D ; therefore the rectangle
contained by A, G is equal to the square of B.

And if the rectangle contained by A, G be equal to the square of
B; A is to B, as BtoG.

The same construction being made, because the rectangle contained
by A, G is equal to the square of B, and the square of B is equal to
the rectangle contained by B, D, because B is equal to D : therefore
the rectangle contained by A, G is equal to that contained by B, D:
but if the rectangle contained by the extremes be equal to that con-
tained by the means, the four stradght lines are proportionals (16.
6.) ; therefore A is to B, as D to G ; but B is equal to D ; wherefore
as A to B, so B to G. Therefore, if three straight lines, &c. Q. K D^

PROP. XVm. PROB.

Ufoit a given straight line to describe a rectilineal figure simi-
lar and similarly situated to a given rectilineal figure.^

Let AB be the given straight line, and GDEF the ^ ven rectilinea}
figure of four sides ; it is required upon the given straight line AB
to deseribe a rectilineal figure sin^ar and similarly situated to
GDEF.

Join DF, and at the points A, B; in the straight fine AB, make (23.
1.) the angle BAG equal to the angle at G, and the angle ABG equal
to the angle GDF; therefore the remainfaig angle GFD is equal to the
remaining angle AGB (32. l.>; wherefore the triangle FGD is equi*

* See Note.
1»

138 THfi BLEMENT8 OF EUCLID* BOOK Tl.

angular to the triangle GAB: H .

again at the points G, B, in G
the straight line GB, make,
(23. 1.) the angle BGH equal
to the angle DFE, and the
angle GBH equal to FDE :
therefore the regaining an-
gle FED is equal to the re- A B CD
maining angle GHB ; and

the triangle FDE equiangular to the triangle GBH : then, because
the angle AGB is equal to the angle CFD, and BGH to DFE, the
whole angle AGH is equal to the whole CFE: for the same reason,
the angle ABH is equal to the angle CDE ; also the angle at A is
equal to the angle at C, and the angle GHB to FED ; therefore the
rectilineal figure ABHG is equiangular to CDEF : but likewise these
figures have their sides about the equal angles proportionals ; be-
cause the triangles GAB, FCD being equiangular, BA is (4. 6.) to
AG, as DC to CF ; and because AG is to GB, as CF to FD j and as
GB to GH, so, by reason of the equiangular triangles BGH, DFE, is
FD to FE; therefore, ex aequali (22. 5.), AG is to GH, as CF to FE :
in the same manner it may be proved that AB is to BH, as CD to
DE : and GH is to HB, as FE to ED (4. 6.). Wherefore, because
the rectilineal figures ABHG, CDEF are equiangular, and have their
sides about the equal angles proportionals, they are similar to one
another (1. def. 6.).

Next, let it be required to describe upon a given straight line AB,
a rectilineal figure similar and similarly situated to the rectilineal
figure CDKEF.

Join DE, and upon the given straight line AB describe the recti«
lineal figure ABHG similar and similarly situated to the quadrilate-
ral figure CDEF, by the former case ; and at the points B, H, in the
straight line BH, make the angle HBL equal to the angle EDK, and
the angle BHL equal to the angle DEK ; therefore the remaining
angle at K is equal to the remaining angle at L : and because the
figures ABHG, CDEF are similar, the angle GHB is equal to the
angle FED, and BHL is equal to DEK ; wherefore the whole angle
GHL is equal to the whole angle FEK : for the same reason the an-
gle ABL is equal to the angle CDK ; therefore the five sided figures
AGHLB, CFEKD are equiangular ; and because the figures AGHB,
CFED are similar, GH is to HB, as FE to ED ; and as HB to HL,
so is ED to EE (4. 6.) ; therefore, ex xquaU (22* 5.)> GH is to HL,
as FE to EK : for the same reason, AB is to BL, as CD to DK : and
BL, is to LH, as (4. 6.) DK to liE ; because the triangles BLH, DKM
are equiangular ; therefore, because the five sided figures AGrHLB,
CFEKD are equiangular, and have their sides about the equal an-
gles proportionals, they are similar to one another ; and in the same
manner a rectilineal figure pf six or more sides may be described
upon a given straight lii^ sixn^ar to one gtfen, and so on. Whieh
was to be done.

BOOK VK THB BLBMBKTS OF BUGUD. 1S9

PROP. XIX. THEOR.

Similar triangles are to one another in the duplicate ratio of
their homologous sides.

Let ABC, DEF be similar triangles, having the angle B equal to
the angle E, and let AB be to BD, as DE to EF, so that the side BC
is homologous to EF (12. def. 5.) ; the triangle ABC has to the trf-
angle DEF the duplicate ratio of that which BC has to EF.

Take BG a third proportional to BC, EF (1 1. 6.), so that BC is to
EF, as PF to BG, and join GA ; then, because as AB to*BC, so DE
to EF, alternately (16. 6.), AB is to DE, as BC to EF: but as BC to"
EF ; so is EF to BG ; therefore (11. 5.), as AB to DE, so is EF to
BG ; wherefore the sides of the triangles ABG, DEF which are about
the equal angles, are reciprocally proportional : but triangles which
have the sides about two equal angles reciprocally proportional, are
equal to one another (15. A

6.): therefore the triangle ^ .^ *

ABG is equal to the triangle ^

DEF: and because as BC
is to EF, so EF to BG ;
and that if three straight

lines be proportionals, the /^ / \ y

first is said (10. def. 5.) to ^ q C £ F

have to the third the dupli-
cate ratio of that which it has to the second ; BC therefore has to
BG the duplicate ratio of that which BC has to EF : but as BC to
BG, so is (1. 6.) the triangle ABC to the triangle ABG. Therefore
the triangles ABC has to the triangle ABG the duplicate ratio of that
which BC has to EF : but the triangle ABG is equal to the triangle
DEF : wherefore also the triangle ABC has to the triangle DEF the
duplicate ratio of that which BC has to EF. Therefore, similar tri-
angles, <Sz;c. Q^ E. D.

CoR. From this it is manifest, that if three straight lines be pro-
portionals, as the first is to the third, so is any triangle upon the first
to a similar and similarly described triangle upon the. second.

PROP. XX. THEOR.

Similar polygons may be divided into the same number of
similar triangles, having the same ratio to one another that the
polygons have : and the polygons have to one another the dupli-
cate ratio of that which their homologous sides have.

Let ABCDE, FGHKL be similar polygons, and let AB be the ho-
mologous side to FG ; the polygons ABCDE, FGHKL may be divi-
ded into the same number of similar triangles, whereof each to each
has the same ratio which the polygons have; and the polygon ABCDE

140 TtIB ELEMENTS OF EUCLID, BOOK TU

has to the polygon FGHEL the duplicate ratio of that which the side
AB has to the side FG.

Join BE, EC, GL, LH : and because the polygon ABCDE is simi-
lar to the polygon FGHKL, the angle BAE is equal to the angle
GFL (1. def. 6.), and BA is to AE, as GF to FL (1. def. 6.) ; where-
fore because the triangles ABE, FGL, have an angle in one equal
to £in angle in the other, and their' sides about these equal angles
proportionals ; the triangle ABE is equiangular (6. 6.)« and therefore
similar to the triangle FGL (4. 6.) ; wherefore the angle ABE is equal
to the angle FGL : and, because the polygons are similar, the whole
angle ABC is equal (1. def. 6.) to the whole angle FGH; therefore
.the remaining angle EBC is equal to the remaining angle LGH : and
because the triangles ABE, FGL are similar, EB is to BA, as LG to
GF (1. def. 6.) ; and also, because the polygons are similar, AB is to
BC, as FG to GH (1. def. 6.) ; therefore ex mquali (22. 5.), EB is to
BC, as LG to GH; that is, the sides about the equal angles EBC,
LGH are proportionals ; therefore (22. 5.) the triangle EBC is equi-
angular to the tri- A M
angle LGH, and si-
milar to It (4. 6.).
For the same rea- E
son, the triangle
ECD likewise is si-
milar to the tri-
angle LKH; there-
fore the similar po- DC K H
lygons ABCDE,
FGHKL are divided into the same number of similar triangles.

Also these triangles have, each to each, the same ratio which the
polygons have to one another, the antecedents being ABE, EBC,
ECD, and the consequents FGL, LGH, LHK : and the polygon
ABCDE has to the polygon FGHKL the duplicate ratio of that which
the side AB has to the homologous side FG.

Because the triangle ABE is similar to the triangle FGL, ABE, has
to FGL the duplicate ratio (19. 6.) of that which the side BE has to
the side GL ; for the same reason, the triangle BEC has to GLH the
duplicate ratio of that which BE has to GL ; therefore, as the trian-
^gle ABE to the triangle FGL, so (11. 5.) is the triangle BEC to the
triangle GLH. Again, because the triangle EBC is similar to the
triangle LGH, EBC has to LGH the duplicate ratio of that which the
side EC has to the side LH : for the same reason, the triangle ECD
has to the triangle LHK, the duplicate ratio of that which EC has
to LH; as therefore the triangle EBC to the triangle LGH, so is (11.
6.) the triangle ECD to the triangle LHK ; but it has been proved
that the triangle EBC is likewise to the triangle LGH, as the triangle
ABE to the triangle FGL. Therefore as the triangle ABE is to the
triangle FGL, so is triangle EBC to triangle LGH, and triangle BCD
to triangle LHK : and therefore as one of the antecedents to one of
the consequents, so are all the antecedents to all the consequents
(12. 5.). Wherefore as the triangle ABE to the triangle FGL, so is
the polygon ABCDE to the polygon FGHKL ; but the triangle ABB

BOOK TU TBIB B^BlfSMTS OP EUCI.fD. 141

has to the triangle F6L, the dufdicate ratio of that which the side AB
has to the homologous side FG. Therefore also the polygon ABCDB
has to the polygon FGHKL the duplicate ratio of that which AB has
to the homologous side FG. Wherefore, similar polygons, Ac. Q.

E. n.

Cor. 1. In like manner, it may be proved, that sunilar four sided
figures, or of any number of sides, are one to another in the duplicate
ratio of their homologous sides, and it has already been proved in
triangles. Therefore, universally, similar rectilineal figures are to
one another in the duplicate ratio of their homologous sides.

CoR. 2. And if to AB, FG, two of the homologous sides, a third
proportional M be taken AB has (10. def 5,) to M the duj^cate ratio g[
that which AB has to FG : but the four sided figure or polygon upon
AB, has to the four sided ^ure or polygon upon FG likewise the
duplicate ratio of that which AB has to FG ; therefore, as AB is to M,
so is the figure upon AB to the figure upon FG, which was also
proved in triangles (Cor. 19. 6.). Therefore, universally, it is mani-
fest, that if three straight lines be proportionals, as the ihrst is to the
third, so is any rectilineal figure upon the first to a similar and simi-
larly described rectilineal figure upon the second.

PROP. XXI. THEOR.

Rectilineal figures which are similar to the same rectilineal
figures, are also similar to one another.

Let each of the rectilineal figures. A, B be similar to the rectilineal
figure C ; the figure A is similar to the figure B.

Because A is similar to C ; they are equiangular, and also have
their sides about the equal angles proportionals (I. def. 6.). Again,
because B is similar to C,
they are equiangular, and
have theur sides about the
equal angles proportionals (1.
def. 6.) : therefore the figures
A, B are each of them equi-
angular to C, and have the
sides about the equal angles of each of them and of C proportionals.
Wherefore the rectilineal figures A and B are equiangular (1. Ax. l.)f
and have their sides about the equal angles proportionals (11. 6.).
Therefore A is similar (I. def. 6.) to B. Q. E. D.

PROP. XXII. THEOR.

If four straight lines be proportionals, the similar rectilineal
figures similarly described upon them shall also be proportionals ;
and if the similar rectilineal figures similarly described upon

149 THfi BLSMEHT8 OF fiVCUO. BOOK TI.

four Straight lines be proportionals^ those straight lines shall be
proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz.
AB to CD, as EF to GH, and upon AB, CD let the similar rectilineal
figures KAB, LCD be similarly described ; and upon EF, GH the
similar rectilineal figures MF, NH in like manner: the rectilineal
figure KAB is to LCD, as MF to NH.

To AB, CD take a third proportional (11. 6.) X; and to EF, GH
a third proportional O : and because AB is to CD, as £«F to GH, and
that CD is (11. 6.) to X, as GH to O; wherefore, ex mqUali (22. 5.),
a» AB to X, so is EF to O : but as AB to X, so is (2 Cor. 20. 6.) the
rectilineal KAB to the rectilineal LCD, and as EF to O, so is (2 Cor.
20. 6.) the rectilineal MF to the rectilineal NH : therefore, as KAB to
LCD, so (11. 5.) is MF to NH.

And if the rectilineal KAB be to LCD, as MF to NH ; tiie straight
line AB is to CD, as EF to GH.

Make (12. 6.) as AB to CD so EF to PR, and upon PR describe
(18« 6.) the rectilineal figure SR similar and similarly situated to

BC D

E F G H PR

either of the figures MF, NH : then, because as AB to CD, so is
EF to PR, and that upon AB, CD are described the similar and
similarly situated rectilineals KAB, LCD, and upon EF, PR, in like
manner, the similar rectilineals MF, SR ; KAB is to LCD, as MF
to SR; but by the hypothesis, KAB is to LCD,.as MF to NH: and
therefore the rectilineal MF having the same ratio to each of the two
NH, SR, these are equal (9. 5.) to one another : they are also similar,
and similarly situated ; therefore GH is equal to PR : and because as
AB to CD, so is EF to PR, and that PR is equal to GH ; AB is to CD,
as EF to GH. If, therefore, four straight lines, &c. Q. E. D.

PROP. XXra. THEOR.

Equiangular paVallelofframs have to one another the ratio
which is compounded of the ratios of their sides.*

Let AC, CF be equiangular parallelograms, having the angle

• See Note.

BOOK VI.. THE SLSMEMT8 OF EUCLID. 148

BCD equal to the angle ECX^ : the ratio of the parallelogram AC to
the parallelogram CF, is the same with the ratio which is compomid-
ed of the ratios of their sides.

Let BG, CG, be placed in a straight line ; therefore DC and CE are
also in a straight line (14. 1.); and complete the parallelogram DG;
and, taku3g any straight line K make (12. 6.) as BC to CG, so K to
L ; and as DC to CE, so make (^2. 6.) L to M : therefore the ratios
of K to L, and L to M, are the same with the ratios of the sides, viz.
of BC to CG, and DC to CE. But the ratio of K to M is that which
is said to be compounded (A. def. 5.) of the ratios of K to L, and L
to M : wherefore a]so K has to M th6 ratio compounded of the ratiod
of the sides; and because as BC to CG, so A D H

Is the parallelogram AC to the parallelo-
gram CH (1. 6.) ; but as DC to CG, so is K
to L; therefore K is (11. 5.) to L, as the »
parallelogram AC to the parallelogram GH :
again, because as DC to CE, so is the pa-
rallelogram CH to the parallelogram CF ;
but as DC to CE, so is L to M ; wherefore
L is (11. 5.) to M, as the parallelogram CH
to the parallelogram CF : therefore since it
has been proved, that as K to L, so is the
pEurallelogram AC to the paralldogram CH ; K L M

and as L to M, so the parallelogram CH to

the parallelogram CF ; ex sequaH (22. 50> K is to M, as the parallelo-
gram AC to the parallelogram CF: but K has to M the ratio
which is compounded of the ratios of the sides ; therefore also the
parallelogram AC has to the parallelogram CF the ratio which is
compounded of the ratios of the' sides. Wherefore, equiangular
parallelograms, &o. Q. E. D.

PROP. XXIV. THEOR.

Thb parallelograms about the diameter of any parallelogram
are similar to the whole and to one another.*

Let ABCD be a parallelogram, of which the diameter is AC; and
EG, HK the parallelograms about the diameter : the parallelograms
EG, HK are similar both to the whole parallelogram ABCD, and to
one another.

Because DC, GF \are parallels, the angle ADC is equal (29. 1.)
to the angle AGF: for the same reason, because BC, EF are pa-
rallels, the angle ABC is equal to the angle AEF : and each of the
angles BCD, EFG is equal to the opposite angle DAB (34. L),
and therefore are equal to one another; wherefore the peurallelo-
grams ABCD, AEFG are equiangular; and .because the angle
ABC is equal to the angle AEF ; and the angle BAC common to
the two triangles BAC, EAF, they are equiangular to one another ;
therefore (4. 6.) $is AB to BC, so is AE to BF: and because the

* See Note.

144

THi; BLEMJBNTS OF EUCLID.

BOOK VI.

opposite sides of parallelograms are A E

equal to one another (34. !.)> AB is
(7. 5.) to AD, as AE to AG ; and DC
to CB, as GF to FE ; and also CD to
DA, as FG to GA ; therefore the sides
of the parallelograms ABCD, AE^G
about the equal angles are proportion-
als; and they are therefore similar ""to
one another (1. def. 6.): for the same
reason, the parallelogram ABCD is similar to the parallelogram
FHCK. Wherefore each of the parallelograms GE, KH is similar
to DB ; but rectilineal figures which are similar to the same recti-
lineal figure are also similar to one another (21. 6.) ; therefore the
parallelogram GE is similar to KH. Wherefore, the parallelograms^
&c. d. E. D.

PROP. XXV. PROB.

To describe a given rectilineal figure which shall be sinrilar
to one, and equal to another given rectilineal figure.*

Let ABC be the given rectilineal figure, to which the figure to be
described is required to be similar, and D that to which it* must be
equ^l. It is required to describe a rectilineal figure similar to ABC,
and equal to D.

Upon the straight line BC describe (Cor. 45. 1.) the parallelogram
BE equal to the figure ABC ; also upon CE describe (Cor. 45. 1.)
the parallelogram CM equal to D, and having the angle FCE eqaal
to the angle CBL ; therefore BC and CF are in a straight line (29.
L 14. 1.), as also LE and EM: betwe^ BC and CF find (13. 6.) a
mean proportional GH, and upon GH describe (18. 6.) the rectilineal
figure KGH similar and similarly situated to the figure ABC ; and
because BC is to GH as GH to CF, and if three straight lines be pro-
portionals, ias the first is to the third, so is (2. Cor. 20. 6.) the figure
upon the first to the similar and similarly described figure upon the
second ; therefore as BC to CF, so is the rectilineal figure ABC to
KGH ; but as BC to CF, so is (1. 6.) the parallelogram BE to the
parallelogram EF : therefore as the rectilineal figure ABC is to KGH,
so is the parallelogram BE to the parallelogram EF (II. 5.) : and the

L EM

rectilineal figiure ABC is equal to the parallelogram BE ; therefore the

"^ See Note.

BOOK VI.

THE BLfiMBllTS OF EUCLID.

146

<, .

rectilineal figure KGH is equal (14. 6.) to the parallelogram EF: but
EF is equal to the figure D ; wherefore also KGH is equal to D ; and
it is similar to ABC. Therefore the rectilineal figure KGH has been
described similar to the figure ABC, and equal to D. Which was to
be done.

PROP. XXVn. THEOR.

If two similar parallelograms have a common angle, and be
similarly situated, they are about the same diameter.

Let the parallelograms ABCD, AEFG be similar and similarly
situated, and have the angle DAB common : ABCD and AEFG are
about the same diameter.

For, if not, let, if possible, the paral- A G D

lelogram BD have its diameter AHC in a
different straight line from AF the diame- j^
ter of the parallelogram EG, and let GF
meet AHC in H ; and through H draw HK E
parallel to AD or BC; therefore the paral-
lelc^ams ABCD, AKHG being about the
same diameter, they are similar to one
another (24. 6.) : wherefore as DA to AB,
so is (1. def. 6.) GA to AK : but because B C

ABCD and AEFG are similar parallelograms, as DA is to AB, so i»
GA to AE ; therefore (II. 5.) as GA to AE, so GA to AK; wherefore
GA has the same ratio to each of the straight lines AE, AK ; and
consequently AK is equal (9. 6.) to AE, the less to the greater, which
is impossible: therefore ABCD and AKHG are not about the same
diameter; wherefore ABCD and AEFG must be about the satoe
diameter. Therefore, if two similar, &c. Q,. E. D.

* To understand the three following propositions more easily, it is
to be observed.

* 1. That a parallelogram is said to be applied to a straight line»
when it is described upon it as one of its sides. Bx. gr. the paral*
lelogram AC is said to be applied to the straight line AB.

* But a parallelogram AE is said to be applied to a straight line AB»
deficient, by a parallelogram, when AD the base of AE is less than
AB, and therefore AE is less than the par- EC G
allelogram AC described upon AB in the
same angle, and between the same parallels,
by the parallelogram DO ; and DC is there-
fore called the defect of AE. A D

* 3. And a paraUelogram AG is said to be applied to a straight line
AB, exceeding by a parallelogram, when AF the base of AG is greater
than AB, and therefore AG exceeds AC the parallelogram described
upon AB in the same angle, and between the same parallels, by the
same parallelogram BG.'

146

THE CLBMBKTS ^P SVCUD.

BOOK* Tt.

PROP. XXVn. THEOR.

Or all parallelograms applied to the same straight line, and
deficient by parallelograms, similar and similarly situated to that
which is described upon the half of the line ; that which is ap-
plied to the half and is similar to its defect, is the greatest.*

Let AB be a straight line divided into equal parts in C; and let the
parallelograra AD be applied to the half AC, which is therefore defi^
cient from the parallelogram upon the whole line AB by the paral-
lelogram CL upon the other half CB : of all the parallelograms applied
to any other parts of AB, and deficient by parallelograms that are
similar and similarly situated to CE : AD is the greatest.

Let AF be any parallelogram applied to AK, any other part of AB
than the half, so as to be deficient from the parallelogram upon .the
whole line AB by the parallelogram KH similar, and similarly situat-
ed to CE. AD is greater than AP.

First, let AK the base of AF, be greater than AC thejialf of AB;
and because CE is similar to the paral- ^ » «

lelogram KH, they are about the same
diameter^(26. 6.) : draw their diameter DB,
and complete the scheme : because the pa-
rallelogram CF is equal (43. 1.) to FE, add
KH to both, therefore the whole CH is equal
to the whole KE: but CH is equal (36. I.)
to CG, because the base AC is equal to the
base CB: therefore CG is equal to KE: to
each of these add CF; then the whole AF
is equal to the gnomon CHL : thereforce CE, or the parallelogram
AD is greater, than the parallelogram AF.
Next, let AK the base of AF be less than
AC, and, the same construction being made,
the parallelogram DH is equal to DO (36.
1.), for HM is equal to MG (34. 1.) because
BC^is equal to C A ; wherefore DH is greater
than LG : but DH is equal (43. L) to DK ;
therefore DK is greater than LG ; to each
of these add AL ; then the whole AD is
greater than the whole AP. Therefore of
all palrallelograms applied, &c. Ct E. D.

G F M

K C

« See Note.

BOOK VL THE BLBMVNT8 OP BUCLID. * 147

PROP, XXVm. PROR

To a given straight line to apply a paraUelogram equal to a
given rectilineal figure, and deficient by a paral^logram similar
to a given parallelogram: but the given rectilineal figure to
vhich the parallelogratn to be applied is to be equal, must not
be greater than the parallelogram applied to half of the given
line, having its defect similar to the defect of that which is to be
applied: that is, to the given parallelogram.*

Let AB be the given straight line, and C the given rectilineal
figure, to which the parallelogram to be applied is required to be
equal, which figure must not be greater than the paraUelogram
applied to the half of the line having its defect from that upon the
whole line similar to the defect of that which is to be applied ; and
let D be the parallelogram to which this defect is required to be
similar. It is required to apply a parallelogram to the straight line
AB, which shall be equal to H G O F

the figure C, and be deficient
from the parallelogram upon
the whole line by a parallelo-
gram sipQilar to D.
^ Divide AB in two equal
parts (10. 1.) in the point £,
and upon BB describe the
parallelogram EBFG similar
(18. 6.) and similarly situated
to D, and complete the pa*
rallelogram AG, which must

either be equal to C or greater ^^^__

than it, by the determination: K n

and if AG be equal to C, then what was required is already done:
for, upon the straight line AB, the parallelogram AG is applied
equal to the figure C, and deficient by the parallelogram EP si-
milar to D : but, if AG be not equal to C, it is greater than it :
and EP is equal to AG ; therefore BF also is greater than C.
Make (25. 6.) the parallelogram KLMN equal to the excess of EP
above €, and similar and similarly situated to D ; but D is simi-
lar to EF, therefore (21. 6.) also KM is similar to EF; let KL
be the homologous side to EG, and LM, to GF: and because EF
is equal to G and KM together, EF is greater tlian KM ; therefore
the straight line EG is greater than KL, and GF than LM : make
GX equal to LK, and GO equal to LM, and complete the paral-
lelogram XGrOP : therefore XO is equal and similar to KM ; but
KM is similar to EF ; wherefore also XO is similar to EF, and
therefore XO and EF are about the same diameter <26. 6.); let
GPB be their diameter, and complete the scheme: then because

•Se^Note.

148 THB ELBMIiinV OF IKJGLID. BOOK VI.

EF is equal to O and KM together, and XO a part of the one is equal
to KM a part of the other, the remainder, viz. the gnomon ERO, is
equal to the remainder C : and because OR is equal (34. 1.) to XS,
by adding SR to each, the whole OB is equal to the whole XB : but
XB is equal (36. 1.) to TE, because the base AE is equal to the base
EB; wherefore also TE is equal to OB; add XS to each, then the
whole TS is equal to the whde, viz. to the gnomon ERO : but it has
been proved that the gnomon ERO is equal to C, and therefore also
TS is equal to C. Wherefore the parallelogram TS, equal to the
given rectilineal tgure C is applied to the given straight line AB defi-
cient by the parallelogram SR, similar to the given one D, because
SR is similar to EF (24. 6.). Which was to be done.

PROP. XXIX. PROB.

To a given straight line to apply a parallelogram equal to a
given rectilineal figure^ exceeding by a parallelogram similar to
another given.*

Let AB be the given straight line, and C the given rectilineal figure
to which the paredlelogram to be applied is required to be equal, and
D the paraUelogram to which the excess of the one to be applied
above that upon the given line is required to be similar. It is re-
quired to apply a parallelogram to the given straight line AB, which
shall be equal to the figure, C exceeding by a parallelogram similar
to D.

Divide AB into two equal parts in the point E, and upon EB de-
scribe (18. 6.) the parallelogram EL similar and similarly situated to
D : and jgfiake (25. 6.) the parallelogram GH equal to EL and C to-
gether, and similar and similarly situated to D ; wherefore GH. is
similar to EL (21. 6.); let KH be the side homologous to FL, and
KG to FE; and because the parallelogram GH is greater than EL,
therefore the side KH is greater than FL and KG than FE : produce
FL and FE, and make FLM equal to KH, and FEN to KG, and
complete the parallelogram MN. MN is therefore equal and simi-
lar to GH, but GH is K H
similar to EL; where-
fore MN is similar to
EL and consequently
EL and MN are about
the same diameter (26.
6.): draw their diame-
ter FX, and complete
the scheme. Therefore,
since GH is equal to
BL and O together,
and that GH is equal
to MN; MN is equal

* See Note.

BOOK ▼!.

THG BUSMSNTS Of EUCLID.

149

to EL and C : take away the common part EL : then tl^e remainder,
viz. the gnomon NOL, is equal to C. And because AE is equal to
BB, the parallelogram AN is equal (36. 1.) to the parallelogram NB,
that is, to BM (43. 1.). Add NO to each ; therefore the whole, viz.
the parallelc^ram AX is equal to the gnomon NOL. But the gno-
mon NOL is equal to C ; therefore also AX is equal to C. Where-
fore to the straight line AB there is applied the parallelogram AX
eqind to the given rectilineal C^ ^ceeding by the parallelogram PO,
which is similar to D, because PO is similar to EL (24. 6.). Which
was to be done.

PROP. XXX. PROB.

To cut a given straight line in extreme and mean ratio.

Let AB be the given straight line ; it is required to cut it in ex*
treme and mean ratio.

Upon AB describe (46. 1.) the square BC, and to AC apply the
parallelogram CD equal to BC, exceeding by the figure AD similar
to BC (29. 6.) : but BC is a square, therefore
also AD is a square ; and because BC is equal
to CD, by taking the common part CE from
each, the remainder BF is equal to the remain- A
der AD: and these figures are equiangular,
therefore their sides about the equal angles
are reciprocally proportional (14. 6.); where*
fore as FE to ED, so AE to EB ; but FJB is
equal to AC (34. i.), that is, to AB ; and ED is
equal to AE : therefore as BA to AE, so is AE
to EB: but AB is greater than AE; where-
fore AE is greater than EB (14. 5.) : therefore the straight line AB is
cut in extreme and mean ratio in E (3. def. 6.). Which was to be
done.

Otherunse.

Let AB be the given straight line ; it is required to cut it in ex-
treme and mean ratio.

Divide AB in the point C, so that the rectangle contained by AB,

BC be equal to the square of AC (11. 2.), Then, 1

because the rectangle AB, BC is equal to the square A C B

of AC, as BA to AC, so is AC to CB (17. 6.) : therefore AB is cut in
extreme and mean ratio in C (3. def. 6.). Which was to be done.

PROP. XXXI. THEOR.

In right angled triangles, the rectilineal figure described upon
the side opposite to the right anele, is equal to the similar and
similarly described figures upon me sides containing the right
angle.*

« See Note.

160 THB BLElCiillTS OF EDC|«U>« BQOK Vf.

Let ABC be a right angled tiiaDgle» having the r^t angle BAG 9
the rectilineal figure described upon BC is equal to the fiiimilar and
similarly described fijs^ures upon BA^ AC.

Draw the p^rpendicuIar AD ; therefore, because in the right angled
triangle ABC, AD is drawn from the right angle at A perpendicular
to the base BC, the triangles ABD, ADC are similar to the whole tri-
angle ABC, and to one another (8. 60« and because tbe triangle ABC
is similar to ADB, as CB to BA, so is BA to BD (4. 6.) ; and becau^
these three straight lines are proportionals, as the first to the third,
so is the figure upon the first to the similar and similarly described
figure upon the second (2 Cor.): therefore as CB to BD, so is the
figure upon CD to the similar and
similarly described figure upon BA :
and, inversely (B. 6.)} as DB to BC,
so is the figure upon BA to that upon
BC ; tor the same reason, as DC to
CB, so is the figure upon CA to that
upon CB. Wherefore as BD and DC -n
together to BC, so are the figures
upon BA, AC to that upon BC (24, 5.) :
but BD and DC together are equal to BC. Therefore the figure de-
scribed on BC is equal (A. 5.) to the similar and similarly described
figures on BA, AC. Wherefore, in right angled triangles, &c. €t E. D.

PROP. XXXIL THEOR. •

If two triangles which have two sides of the one proportional
to two sides of the other, be joined at one angle, so as to have
their homologous sides parallel to one another, the remaining
sides shall be in a straignt line.^

Let ABC, DCE be two triangles which have the two sides BA, AC
proportional to the two CD, DE, viz. BA to AC, as CD to DE ; and
let AB be parallel to DC, as AC to DE. BC and CE are in a straight
line.

Because AB is parallel to DC,
and the straight line AC meets
ti)em, the alternate angles BAC,
ACD are equal (29. 1.) ; for the
same reason, the angle CDE is
equal to the angle ACD ; where-
fore also BAC is equal to CDE :
and because the triangles ABC^ B C E

DCE have one angle at A equal to one at D, a^d the sides about
these angles proportionals, viz. BA to AC, as CD to DE, the
triangle ABC is equiangular (6. 6.) to DCE: therefore the angle
AfiC is equal to the angle DCE ; and the angle BAC was proved
to be equal to ACD; therefore the whole angle ACE is equal to

, I

BOOK ▼!. THB BtifiltEllTB OF EOCLip. 151

the two angles ABC, BAC; add ihe eominon angle ACB, then the
angles ACE> ACB are equal to the angles ABC, BAC, ACB : but
ABC, BAC, ACB are equal to two right angles (32. 1.) ; therefore
also the angles ACE, ACB are equal to two right angles : and since
at the point C, in the straight line AC, the tWo straight lines BC, CE,
which are on the opposite sides of it, make the adjacent angles ACE,
ACB equal to two right angles ; therefore (14. 1.) BC and CE are in
a straight line. Wherefore, if two triangles, &c. CI. K D.

PROP. XXXni. THBOR.

Is equal circles, angles, whether at the centres or circumfe-
rences, have the same ratio which the circumferences on which
they stand have to one another : so also have the sectors.*

Let ABC, DEF be equal circles : and at their centres the angles
BGC, EHF, and the angles BAD, E£>F at their circumferences : as
the circumference BC to the circumference EF, so is the angle BGC
to the angle EHF, and the angle BAC to the angle EDF ; and also
the sector BGC to the sector EHF.

Take any number of circumferences CK, KL, each equal to BC,
and any number whatever FM, MN, each equal to EF : and join
GK, GL, HM, HN. Because tlie circumferences BC, CK, KL are all
equal, the angles BGC, CGK, KGL are also all equal (27. 3.) : there^
fore what multiple soever the circumference BL is of the drcum*
ference BC, the same multiple is the angle BGL of the angle BGC :
for the same reason, whatever multiple of the circumference EN is of
the circumference EF, the same multiple is the angle EHN of the
angle EHF: and if the circumference BL be equal to the circumfe-
rence EN, the angle BGL is also equal (27. 30 to the angle EHN ;
and if the circumference BL be greater than EN, likewise the angle
BGL is greater than EHN ; and if less, less: there being then four
magnitudes, the two circumferences BC, £1F, and the two angles
BGC, EHF : of the circumference BC, and of the angle BGC, have
been taken any equimultiples whatever, viz. the circumference BL,
and the angle BGL ; and of the circumference EF, and of the angle
EHF, any equimultiples whatever, viz. the circumference EN, and

* See Note.

THE ELEMENTS OF £Uq|<lD.

BOQ^; vt*

the angle £HN : and it has been proved that if the circumference BL
be greater than EN, the aagle BGL is greater than £HN : and if
equal, equal ; and if less, less : as therefore the circumference BC to
the circumference £F, so (5. def 5.) is the angle BGC to the angle
EHF : but as the angle BGC is to the angle EHF, so is (15. 5.) the
angle BAG to the angle EDF, for each is double of each (20. 3.) :
therefore, as the circumference BC is to £F, so is the angle BGfC to
the angle EHF, and the angle BAG to the €mgle EDF.

Also, as the circumference BC to EF, so is the sector BGG to the
sector EHF. Join BC, GEl, and in the circumferences BC, GK take
any points X, O, and join BX, XC, CO, OK : then, because in the
triangles GBG, GCK, the two sides BG, GG are equal to the two
CG, GK, and that they contain equal angles ; the base BG is equal
(4. 1.) to the base CK, and the triangle GBC to the triangle GGK :
and because the circumference BC is equal to the circumference CK,
the remaining part of the whole circumference of the circle ABC, is
equal to the remaining part of the whole circumference of the same
circle : wherefore the angle BXC is equal to the angle COK (27..
3.), and the segment BXC is therefore similar to the segment COK
(11. def 3.): and they are upon equal straight lines BC, CK: but
similar segments of circles upon equal straight lines, are equal
(24. 3.) to one another: therefore the segment BXC is equal to
the segment COK: and the triangle BGC is equal to the triangle
CGK ; therefore the whole, the sector BGC, is equal to the whole,
the sector CGK: for the same reason, the sector KGL is equal to
each of the sectors BGC, CGK : in the same manner, the sectors
EHF, FHM, MHN may be proved equal to one another: there-
fore, what multiple soever the circumference BL is of the circum*
ference BC, the same midtiple is the sector BGL of the sector
BGC: for the same reason, whatever multiple the circumfeernce
EN is of EF, the same multiple is the sector EHN of the sector
EHF : and if the circumference BL be equal to EN, the sector BGL

is equal to the sector EHN : and if the circumference BL be greater
than EN, the sector BGL is greater than the sector EHN ; and if
less, less : since^ then, there are four magnitudes, the two chrcum-
ferences BC, EF, and the two sectors BGC, EHF, and of the circum-
ference BC, and sector BGC, the curcumference BL and sector BGL
are any equt^l multiples whatever: and of the circumference EF, and

BOOK VI. THC SLBMBirrs OF EUCLIIX 15®

sector MIF, the Gircuml^rence EN and sector EHN are any eq;Qi*
multiples whatever; and that it has been proved, if the circumference
BL be greater than EN, the sector BGL is greater than the sectcnr.
EHN ; and if equal, equal ; and if less, less: Therefore (6. def. 5.)i a«
the circumference BC is to the circumference EF, so is the sector
B6C to the sector EHF. Wherefore, in equal circles, ^. Q. E. D.

PROP. B. THEOR.

If an angle of a triangle be bisected by a straight line, which
likewise cuts the base ; the rectangle contained by the sides of the
triangle is equal to the rectangle contained by the segments of
the base, together with the square of the straight line bisecting
the angle.*

Let ABC be a triangle, and let the angle BAG be Wsected by the
straight line AD ; the rectangle BA, a6 is equal to the rectangle BD,
DC together with the square of AD.

Describe the circle (5. 4.) ACB about the triangle, and produce AD
to the circumference in E, and join EC :
then because the angle BAD is equal to
the angle CAE, and the angle ABD to
the angle (21. 3.) AEC, for they are in

the same segment : the triangles ABD, g j^ f M q

AEC, are equiangular to one another :
therefore as BA to AD, so is (4. 6.) EA
to AC, and consequently the rectangle
BA, AC is equal (16. 6.) to the rectan-
gle EA, AD, that is (3. 2.), to the rect-
angle ED, DA, together with the square
of AD : but the rectangle ED, DA, is
equal to the rectangle (35. 3.) BD, DC. Therefore the rectangle BA,
AC is equal to the rectangle BD, DC, together with the square of AD.
Wherefore, if an angle, &c. Q. E. D.

PROF. C. THEOR.

If from any angle of a triangle a straight line be drawn per-
pendicular to the base ; the rectangle contained by the sides ol the
triangle is equal to the rectangle contained by the perpendicular
and the diameter of the circle described about the triangle.*

Let ABC be a triangle, and AD the perpendicular from the angle
A to the base BC ; the rectangle BA, AC is equal to the rectangle
contained by AD and the diameter of the circle described about the
triangle.

* See NotCe
20

154

THB SLEMENTS OF EUCLID.

BOOK VI.

Describe (5. 4.) the circle AOB about
the triangle, and draw its diameter AB,
and join EC; because the right angle
BDA is equal (31. 8.) to the angle EGA
in a semicircle, and the angle ABD to B
the angle AEC in the same segment (21. i
3.) ; the triangles ABD, AEC are equi-
angular : therefore as (4. 6.) BA to AD,
so is EA to AC ; and consequently the
rectangle BA, AC is equal (16. 6.) to the
rectangle EA, AD. If, therefore, from
an angle, &c. Qt. E. D.

PROP. D. THEOR.

The rectangle contained by the diagonals of a quadrilateral
inscribed in a circle, is equal to both the rectangles contained by
its opposite sides.*'

Let ABCD be any quadrilateral inscribed in a circle, and join AC,
BD ; the. rectangle contained by AC, BD is equal to the two rectan-
gles contahied by AB, CD, and by AD, BC.t

Make the angle ABE equal to the angle DBC ; add to each of these
the common angle EBD, then the cmgle ABD is equal to the angle
EBC; and the angle BDA is equal (21. 3.) to the angle BCE, be-
cause they are in the same segment ; therefore the triangle ABD is
equiangular to the triangle BCE ; where- B
fore (4. 6.) as BC is to CE, so is BD to
DA; and consequently the rectangle
BC, AD is equal (16. 6.) to the rectan-
gle BD, CE : again, because the angle
ABE is equal to the angle DBC, and
the angle (21. 3.) BAE to the angle
BDC, the triangle ABE is equiangular
to the triangle BCD : as therefore BA
to AE, so is BD to DC ; wherefore the

rectangle BA, DC is equal to the rectan-

gle BD, AE : but the rectangle BC, AD has been shown equal to the
rectangle BD, CE; therefore the whole rectangle AC, BD (1. 2.) is
equal to the rectangle AB, DC, together with the rectangle AD, BC.
Therefore the rectangle, &c. ft. E. D.

* See Note.

t Tbii is a Lemma of CI. Ptolofflens, in page 9 of his fAry^LKu «-uFr«(is.

THE

ELEMENTS OF EUCLID.

BOOK XL

DEFINITIONS.

L
A SOLID is that which hath length, breadth and thiekneM.

That which bounds a solid is a superficies.

ra.

A straight line is perpendicular, or at right angles to a plane, when
it makes right angles with every straight line meeting it in that
plane.

IV.

A plane is perpendicular to a plane, when the straight lines drawn
in one of the planes perpendicularly to the common section of the
two planes, are perpendicular to the other plane.

■ ^ ^•

The inclination of a straight line to a plane is the acute angle con-
tained by that straight line, and another drawn from the point in
which the first line meets the plane to the point in which a perpen^
dicular to the plane drawn from any point of the first line above
the plane, meets the same plane.

VI.

The inclination of a plane to a plane as the acute angle contained
by two straight lines drawn from any the same point of their
common section at right angles to it, one upon one plane, and the
other upon the other plane.

vn.

Two planes are said to have the same, or a like inclination to one
another, which two other planes have, when the said angles of
inclination are equal to one another.

186 THB fiLfiMSMTS OP EUCLID. BOOK. XI.

vm.

Parallel planes are such which do not meet one another though pro-
duced.

IX.

A solid angle is that which is made by the meeting of more than two
plane angles, which are not in the same plane, in one point.*

Jv..

* The tenth definition is omitted for reasons given in the notes.''^

Similar solid figures are such as have all their solid angles equal,
each to each, and which are contained by the same number of
similar planes.* '

XII.

A pyramid is a solid figure contained by planes that are constituted
betwixt one plane and one point above it in which they meet.

Xffl.
A prism is a solid figure contained by plane figures, of which two
that are opposite are equal, similar, and parallel to one another ;
and the others parallelograms.

XIV.
A sphere is a solid figure described by the revolution of a semicircle
about its diameter, which remains unmoved.

XV.

The axis of a sphere is the fixed straight line about which the s^xii-
circle revolves.

XVI.
The centre of a sphere is the same with that of a semicircle.

xvn.

The diameter of a sphere is any straight line which passes through
the centre, and is terminated both ways by the superficies of the
sphere.

xvm.

A cone is a solid figure desicribed by the revolution of a right angled
triangle about one of the sides containing the right angle, which
side remains fixed.

If the fixed side be equal to the other side containing the right angle,
the cone is called a right angled cone ; if it be less than the other
side, an obtuse angled, and if greater, an acute angled con^.

XIX.

The axis of a cone is the fixed straight line about which the triangle
revolves.

XX.
The base of a cone is the circle described by that side containing
the right angle, which revolves.

XXL
A cylinder is a solid figure described by the revolution of a right an-
gled parallelogram about one of its sides which remains fixed.

• See Note.

BOOK XI. THE ELEMISNTS OF ttVCUD. 4^7

XXII.

The axis of a cylinder is the fixed straight line about which the
parallelogram revolves.

xxm.

The bases of a cylinder are the circles described by the two revolv-
ing opposite sides of the parallelogram.

XXIV.

Similar cones and cylinders are those which have their axes and the
diameters of their bases proportionals.

XXV.

A cube is a solid figure contained by six equal squares.

XXVI.

A tetrahedron is a solid figure contained by four equal and equila-
teral triangles.

xxvn.

An octahedron is a solid figure contained by ei^t equal and equi-
lateral triangles.

xxvm.

A dodecahedron is a solid figure contained by twelve equal penta-
gons which are equilateral and equiangular.

XXIX.
An icosahedron is a solid figure contained by twenty equal and equi-
lateral triangles.

DEF. A,
A parallelopiped is a solid figure contained by six quadrilateral figures,
whereof every opposite two are parallel.

PROP. L THEOR.

One part of a straight line cannot be in a plane, and another
part above it.*

If it be possiWe, let AB, part of the straight Ihie ABC, be m the
plane, and the part BC above it : and since the straight line AB is
in the plane, it can be produced in
that plane : let it be produced to D :
and let any plane pass through the
straight line AD, and be turned
about it until it pass through the

point C : and because the points B, Care in this plane, th^ straight line
BC is in it (7. def. 1.) : therefore there are two straight lines ABC,
ABD in the same plane that have a common segment AB, which is
impossible (Cor. 11. 1.). Therefore, one part, &c. Q. E. D.

PROP. n. THEOR.
Two straight lines which cut one another are in one plane»

•SeeNote.

168

THS BbBMfiWTa OF BUCLID.

BOOK ZI.

and three straight lines which meet one another are in one
plane/

Let two straight lines AB, CD cut one another in E ; AB, CD kre
in one plane : and three straight lines EC, CB, BE which meet one
another, are in one plane.

Let any plane pass through the straight line

EB, and let the plane be turned about EB, pro-
duced, if necessary, untO it pass through the
point C : then because the points E, C are in
this plane, the straight Ime EC is in it (7. def.
1.) : for the same reason, the straight line BC
is in the same ; and, by the hypothesis, EB is
in it ; therefore the three straight lines, EC, CB,
BE are in one plane : but in the plane in which

EC, EB are, in the same are (1. 11.) CD, AB : C " ' ^ B
therefore AB, CD are in one plane. Wherefore two straight linea,
&& Q. £. D.

PROP. m. THEOR.

If two planes cut one another, their common section is a
straight line.*

Let two planes AB, BC cut one another, and let the line DB be
their common section: DB is a 'straight line :
ifitbenot^ from the point D to B, draw, in
the plabe AB, the straight line DEB, and in
the plane BC the straight line DFB ; then two
straight lines DEB, DFB have the same ex-
tremities, and therefore Include a space betwixt
them; which is impossible (10. At, 1.): there-
fore BD the common section of the planes AB,
BC cannot but be a straight line. Wherefore,
if two planes, &c. €1. E. D.

PROP. IV. THEOR.

If a straight line stand at right angles to each of two straight
lines in the point of their intersection, it shall also be at right an-
gles to the plane which passes through them, that is, to the plane
in which they are.*

Let the straight line £F stand at right angles to each of the straight
lines AB, CD in E, the point of their intersection : EF is also at right
angles to the plane passing through ABv CD.

Take the straight lines AE, EB, CE, ED all equal to one
another; and through E draw, in the plane in which are AB,
CD, any straight line GEH ; and join AD, CB : then, from any
point F in EF, draw FA, FG, FD, FC, FH, FB: and beoaose the

«SMN0le.

BOOK XL THB aLBIISHTa OF EDlAID. IB9

two stralgbt Hnes, AB, ED are equal to the two BE, EC, and that
they contain equal angles (IS. 1.) AED, BBC, the base AD Is equal
(4. 1.) to the baae EC, and the angle DAE to the angle EBC: and
the angle AEG is equal to the angle BEH (IB. 1.); therefore the tri-
angles AEG, BEH have two angles of one equal to two angles of the
other, each to each, and the sides AE, EB, adjacent to the eqnal an-
gles, equal to one another; wherefore they shall have their other
sides equal (26. 1.) : GE is therefore equal to BH, and AG to BH:
and because AB la equal to BB, and FB common and at right angles
to them, the base AB is equal (4. 1 .) to the base FB ; for the same
reason, CP is equal to FD: and becauae AD is equal to BC, and AF
to FB, the two sides FA, AD are equal to the two FB, BC, each to
each, and the base DF was proved equal to F

the base BC; therefore the angle FAD is
equal (8. 1.) to the angle FBO : again, it was
proved that GA ia equal to BH, and also AF
to FB ; FA, then, and AG are equal to FB

and BH, and the angle FAG has beeh proved "

equal to the angle FBH ; therefore the base
GF (4. 1.] to the base FH : again, because
it was proved, that GK is equal to BH, and

EF is common ; GE, EF are equal to HE, H

EF; and the base GF is equal to the base
FH ; therefore the angle GBF is equal {8. 1.)
to the angle HEF ; and consequently each
of these angles is a right (10. def 1.) angle. "

Therefore FB makes right angles with GH, that is, with any straight
line drawn through E in the plane passing tlirough AB, CD. In like
manner, it may be proved, that FE makes right angles with every
straight line which meets it in that plane. But a straight line is at
right angles to a plane when it makes right angles with every straight
line which meets it in that plane (3. def 11,): therefore EF is at right
angles to the plane in which are AB, CD. Wherefore, if a straight
line, &.C Q. E. D.

PROP. V. THEOR.

Ir three straight lines meet all in one point, and a straight line
stands at right angles to each of them in that point ; these three
straight lines are in one and the same plane.*

Let the straight line AB stand at right angles to each of the straight
Ihies BC, BD, BE, hi B the point where they meet; BC, BD, BE are
in one and the same plane.

If not let, if it be possible, BD and BE be fn one jrfane, and BC be
above it; and let a plane pass through AB, BC, the common section
of wMdi with the plane, hi which BD and BE are, shall be a straight

• See Note.

THfi ELEBlBllTa OF SUOLID.

BOOS XI.

(3, 11.) line; let this be BF: therefore the three straight lines AB;

BC, BF are all in one plane, viz. that which passes through AB, BC :
and because AB stands at right angles to each of the straight lines

BD, BE, it is also at right angles (4. 11.) to the plane passing through
them; and therefore roakes right angles (3* def. U.) .with every
straight line meeting it in that plane; but
BF, which is in that plane, meets it : there- A
fore the angle ABF is a right angle ; but the
angle ABC, by the hypothesis, is also a right
angle ; therefore the angle ABF is equal to
the angle ABC, and they are both in the
same plane which is impossible: therefore
the straight line BC is not above the plane
in which are BD and BE: wherefore the
three straight lines BC, BD, BE are in one
and the same plane. Therefore, if three straight lines, Sic. Ct> £L D«

PROP* VI. THEOR.

Ir two straight lines be at right angles to the same plane, they
shall be parallel to one another.

Let the straight lines AB, CD be at right angles to the same plane f
AB is parallel to CD. ^

Let them meet the plane in the points B, D, and draw the straight
line B, D, to which draw DE at right angles, in the same plane ; and
make DE equal to AB, and join BE, AE, AD. A
Then because AB is perpendicular to the plane,
it shall make right (3. def. 11.) angles with every
straight line which meets it, and is in that plane:
but BD, BE, which are in that plane, do each
of them meet AB. Therefore, each of the an-
gles ABD, ABE is a right angle : for the same B
reason, each of the angles CDB, CDE is a right
angle : and because AB is equal to DE, and BD
common, the two sides AB, BD are equal to the
two ED, DB : and they contain right angles ;
therefore the base AD is equal (4. I.) to the
base BE: again, because AB is equal to DE, and BE to AD; AB»
BE are equal to ED, DA ; and, in the triangles ABE, EDA, the
base ^E is common ; therefore the angle ABE is equal (8. 1.) to
the angle EDA : but ABE is a right angle ; therefore EDA is also
a right angle, and ED perpendicular to DA: but it is also per-
pendicular to each of the two BD, DC : wherefore ED is at right
angles to each of the three straight lines BD, DA, DC, in the point
in which they meet : therefore these three straight hnes are all in
the ^ame plane (5. 11.):. but AB is in the plane in which are BD,
DA, because any three straight lines which meet one another are
in one plane (2. 11.): therefore AB, BD, DC are in one plane:

BOOK XI. THB SLEMBNT8 OF BUCLID. 191

and each of the angles ABD, BDC is a right angle; therefore AB is
parallel (28. 1.) to CD. Wherefore, if two straight lines, &c CI. E. IX

PROP. Vn. THEOR.

If two straight lines be parallel, the straight line drawn from
any point in the one to any point in the other is in the same plane
with the parallels.*

Let AB, CD be parallel straight lines, and take any point E in. the
one, and the point F in the other*: the straight Ihie which joins E
and F is in the same plane with the parallels.

If not, let it be, if possible, above the plane, as EGF ; ^and in the
plane ABCD in which the parallels A E B

are, draw the straight line EHF from
£ to F; and since EGF also is a
straight line, the two straight lines
EHF, EGF includes a space between
them, which is impossible (10. Ax. 1.).
Therefore the straight line joining the
points E, F is not above the pkme in
which the parallels AB, CD are, and is therefore in tiiat plane. Where^
fore, if two straight lines, d^c. Q. E. D.

PROP. Vm. THEOR.

It two straight lines be parallel, and one of them be at right
angles to a plane, the other also shall be at right angles to the
same plane.*

Let AB, CD be two parallel straight lines, and let one of them AB
be at right angles to a planer the other CD is at right angles to the
same plane.

Let AB, CD meet the plane in the points B, D, and join BD:
therefore (7. 11.) AB, CD, BD are in one plane. In the plane ta
which AB is at right angles, draw DE at right angles to BD, and
make DE equal to AB, and join BE, AE, AD. And because AB is
perpendicular to the plane, it is perpendicular to every straight
line which meets it, and is in that plane (8. def. II.); therefore
each of the angles ABD, ABE is a right angle; and because
the straight line BD meets the parallel straight lines AB, OD,
the angles ABD, CDB are together equal (29. 1.) to two right
angles: and ABD is a right angle; therefore also CDB is a right
angle, and CD perpendicular to BD: and because AB is. equal
to DE, and BD common, the two AB, BD are equal to the two ED,
DB, and the angle ADB ise^al to the angle EDB> because each of

» See Note*
21

162

THE ELEMSMTS OF fiUCLU).

BOOK XI.

them is a right an^e ; therefore the base AD
is equal (4. 1.) to the base BE: again, because A
AB is equal to D£, and EB to AD ; the two
A6, BE are equal to the two ED, DA ; and
the base AE is commcm to the triangles ABE,
EDA; wherefore the angle ABE is equal (8. 1.)
to the angle EDA, and ABE is a right angle ; ,
and therefore EDA, is a right angle, and ED
perpendicular to DA : but it is also perpendicu- B
lar to BD: therefore ED is perpendicular
(4. 11.) to the plane which passes through
BD, DA, and shall (3. def. 11.) make right an-
gles with every straight line meeting it in that
plane : but DC is in the plane passing through
BD, DA, because all three are in the plane in
which are the parallels AB, CD; wherefore ED is at right angles to
DC ; and therefore CD is at right angles to DE : but CD is also at
right angles to DB; CD then is at right angles to the two straight
lines DE, BD in the point of their intersection D; and therefore is at
right angles (4. 11.) to the plane passing through DE, DB, which is
the same plane to which AB is at right angles. Therefore, if two
straight lines, &c. Q,. E. D.

PROP. IX. THEOR.

Two straight lines which are each of them parallel to the same
straight line, and not in the same plane with it, are parallel to
one another.

Let AB, CD be each of them parallel to EF, and not in the same
plane with it ; AB shall be parallel to CD.

In EF take any point G, from which draw, in the plane passing
through EP, AB, the straight line GH at right apgles to EF, and in
the plane passing through EF, CD, draw GK at right angles to the
same EF. And because EF is perpendicular both to GH and GK,
EF is perpendicular (4. 11.) to the A H B

plane HGE passing through them: and
EF is parallel to AB, therefore AB is at
right angles (8. 11.) to the plane HGK.
For the same reason, CD is likewise
at right angles to the {Jane HGK.
Therefore, AB, CD are each of them
at right angles to the plane HGK. But ^
if two straight lines be at right angles
to the same plane, they shall be parallel (a 11.) to one another.
Therefore AB is parallel to CD. Wherefore, two straight lines, &c.
Q. E.D.

PROP. X. THEOR.

If two straight lines meeting one another be parallel to two
other that meet one another, and are not in the same plane

BOOK XL

THB JBLSMBMTB OF RUCLID.

163

with the first two, the first two and the other t^o shall contain
equal angles.*

Let the two straight lines AB, BC which meet one another be
parallel to the two straight lines D£, EF that meet one another, and
are not in the same plane with AB, BC. The angle ABC is equal
to the angle DEF.

Take BA, BC, ED, EF all equal to one another ; and join AD, CF,
BE, AC, DF ; because BA is equal and parallel to ED« therefore AD
is (33. 1.) both equal and parallel to BE. For B

the same reason, CF is equal and parallel to
BE. Therefore AD and CF are each of them
equal and parallel to BE. But straight lines A
that are parallel to the same straight line, and
not in the same plane with it, are parallel (9.
11.) to one another. Therefore AD is paral-
lel to CF ; and it is equ^l (1. Ax. 1.) to it, and
AC, DF join them towards the same parts ;
and therefore (33. 1.) AC is equal and parallel
to DF. And because AB» BC are equal to
DE, EF, and the base AC to the base DF ; D
the angle ABC is equal (8. 1.) to the angle DEF. Therefore, if two
straight lines, &c. d. E. D.

\_ C

PROP. XI. PROB.

To draw a straight line perpendicular to a plane, from a given
point above it

Let A be the given point above the plane BH ; it is required to
draw from the point A a straight line perpendicular to the plane BH.

In the plane draw any straight line BC, and from the point A
draw (12. 1.) AD perpendicular to BC. If then AD be also perpen-
dicular to the plane BH, the thing required is already done ; but if
it be not, from the point D draw (11. 1.) in the plane BH, the straight
line DE at right angles to BC : and
from the point A draw AF perpen-
dicular to DE ; and through F draw
(31. 1.) GH parallel to BC : and be-
cause BC is at right angles to ED and
DA : BC is at right angles (4. 11.) to
the plane passing through ED, DA.
And GH is parallel to BC : but, if two
straight lines be parallel^ one of which
is at right angles to a plane, the other B D C

shall be at right (8. 11.) angles to the same plane ; wherefore GH is
at right angles to the plane through ED, DA, and is perpendicular
(3. def. 11.) to every straight line meeting it in that plane. But AF,
which is in the plane through ED, DA meets it: therefore GH is

164 THB BLEMfiNTB OP BUOLID. BOOK Xf.

perpendicular to AF ; and consequently AF is perpendicular to GH,
and AF is perpendicular to DE : therefore AF is perpendicular to
each of the straight lines GH, DK But if a straight line stands at
right angles to each of two straight lines in the point of their inter-
section, it shall also be at right angles to the plane passing through
them. But the plane passing through ED, GH is the plane BH ;
therefore AF is perpendicular to the plane BH ; therefore, from the
given point A, above the plane BA, the straight line AF is drawn
perpendicular to that plane. Which was to be done.

PROP. XII. PROB.

To erect a straight line M right angles to a given plane, from
a point given in the plane.

Let A be the point given in the plane ; it is required to erect a
straight line from the point A at right angles D B

to the plane.

From any point B above the plane draw
(11. 11.) BC perpendicular to it; and from
A draw (31. 1.) AD parallel to BC. Because,

therefore, AD, CB are two parallel straight / [ [~ 7

lines, and one of them BC is at right angles / ^1 ^ /

to the given plane, the other AD is also at / /

right angles to itt(8. 11.). Therefore a

straight line has been erected at right angles to a given plane from

a point given iii it. Which was to be done.

PROP. Xm. THEOR.

From the same point in a given plane, there cannot be two
straight lines at right angles to the plane, upon the same side of
it; and there can be but one perpendicular to a plane frofti a

point above the plane.

•
For, if it be possible, let the two straight lines AC, AB be at right
angles to a given plane from the same point A in the plane, and upon
the same side of it : and let a plane pass through BA, AC ; the com-
mon section of this with the given plane is a straight (3. 11.) line
passing through A : let DAE be their common section : therefore
the straight lines AB, AC, DAE are in one plane : and because CA
is at right angles to the given plane, it shall make riglit angles with
every straight line meeting it in that plane. But DAE, which is in that
plane, meets CA; therefore CAE is a B C

right angle. For the same reason BAE
is a right angle. Wherefore the angle
CAE is equal to the angle BAE ; and
they are in one plane, which is impossi-
ble. Also, from a point above a plane,
there can be but one perpendicular to
that plane ; for, if there could be two,

BOOK XI.

THB BLEMEirrS OF EUCLID.

165

they would be paraUel (6. 11.) to one another, 'which is absurd.
Therefore, from the same point, &c. d. E. D.

PROP. XIV. THEOR.

Plaices to which the same straight line is perpendicular, are
parallel to one another.

Let the straight line AB be perpendicular to each of the planes CD,
£F ; these planes are parallel to one another.

If not, they shall meet one another when produced; let them
meet ; their common section shall be a G

straight line GH, in which take any point
K, and join AE[, BK: then because AB is
perpendicular to the plane EF, it is perpen-
dicular (3. def. 11.) to the straight line BK
which is in that plane. Therefore ABK is
a right angle. For the same reason, B AK
is a right angle ; wherefore the two an-
gles ABE, BAK of the triangle ABK are
equal to two. right angles, which is im-
possible (17. i.) ; therefore the planes CD,
KF though produced, do not meet one
another ; that is, they are parallel (8. def.
1 1.). Therefore, planes, &c. Q. E. D.

PROP. XV. THEOR.

If two straight lines meeting one another, be parallel to two
straight lines which meet one another, but are not in the same
plane with the first two, the plane which passes through these is
parallel to the plane passing the others.*

Let AB, BC, two straight lines meeting one another, be parallel to
DE, EF, that meet one another, but are not in the same plane with
AB, BC : the planes through AB, BC, and DE, EF shall not meet,
though produced.

From the point B draw BG perpendicular (11. 11.) to the plane
which passes through DE, EF, and let it meet that plane in G;
and through G draw GH parallel (31. 1.) to ED, and GK parallel
to EF ; and because BG is perpendicular to the plane through
DE, EF, it shall make right an-
gles with every straight line meet-
ing it in that plane (3. def. 11.).
But the straight lines GH, GK,
in that plane meet it: therefore
each of the angles BGH, BGK is
a right angle : and because BA is
parallel (9. 11.) to GH (for each
of them is parallel to DE, and
they are not both in the same

cT^

"^"^■^^^^^

'^ See Note.

166

TR8 ELEMENTS OP EOCLID.

BOOK XI.

plane with It) the angles GBA, BGH are together equal (29. 1.) to
two right angles : and BGH is a right angle, therefore also GBA is a
right angle, and GB perpendicular to BA ; for the same reason, GB
is perpendicular to BC : since therefore the straight line GB stands
at right angles to the two straight lines BA, BC, that cut one another
inB; GB is perpendicular (4. 11.) to the plane through AB, BC :
and it is perpendicular to the plane through DE, EP : therefore BG
is perpendicular to each of the planes through AB, BC, and DEJ, EP:
but planes to which the same straight line is perpendicular, are pa-
rallel (14. 11.) to one another: therefore the plane through AB, BC is
parallel to the plane through DE, EP. Wherefore, if two straight
lines &c. Q. E. D.

PROP. XVI. THEOR.

If two parallel planes be cut by another plane, their common
sections with it are parallels.*

Let the parallel planes AB, CD be cut by the plane EFHG, and
let their common sections with it be EP, GH; EP is parallel to GH.

For, if it be not, EP, GH shall me^t, if produced, either on the
side of PH, or EG; first, let them be produced on the side of PH,
and meet in the point K ; therefore, since EPK is in the plane AB,
every point in EFK is in that plane; K

and K is a point in EPK ; therefore
^ is in the plane AB : for the same
reason K is also in the plane CD:
wherefore the planes AB, CD pro-
duced meet one another; but they
do not meet, since they are parallel
by the hypothesis; therefore the
straight lines EF, GH do not meet
when produced on the side of PH ;
in the same manner it maybe proved,
that EP, GH do not meet when pro-
duced on the side of EG : but straight lines which are in the same
plane and do hot meet, though produced either way, are parallel :
therefore EP is parallel to GH. Wherefore, if two parallel planes,
&.C. Q. E. D. /

PROP. XVn. THEOR.

Ir two straight lines be cut by parallel planes, they shall be cut
in the same ratio.

Let the straight lines AB, CD be cut by the parallel planes GH,
KL. MN, in the points A, E, B; C, F, D; as AE is to EB, so is CP
toFD.

Join AC, BD, AD, and let AD meet the plane KL in the
point X ; and join EX, XF : because the two parallel planes

*SMNote.

BOOK Xh

TBfi fiLBMBMTS OP EUCLID.

167

-/H

KL, MN are cut by the plane EBDX, the common sections! EX*
BD, are parallel (16. 11.). For the same reason, because the two
parallel j^anes GH, KL are cut by
the plane AXFC, the common
sections AC, XF, are parallel:
and because EX is ' parallel to
BD, a side of the triangle ABD,
as A£ to EB, so is (2. 6.) AX to
XD. Again, because XF is pa-
rallel to AC, a side of the triangle
ABC, as AX to XD, so is CF to
¥D : and it was proved that AX
is to XD, as AE to EB; there-
fore (IL 5.), as AE to EB, so is
CF to FD. Wherefore, if two
straight lines, &c. Q. E. D.

PROP. XVni. THEOR.

If a straight line be at right angles to a plane, every plane
-which passes through it shall be at right angles to that plane.

Let the straight line AB be at right angles to a plane CK ; every
plane which passes through AB shall be at right angles to the plane
CK.

Let any plane DE pass through AB, and let CE be the common
section of the planes DE, CK; take any point F in CE, from which

draw FG in the plane DE at right
angles to CE ; and because AB is
perpendicular to the plane CK,
therefore it is also perpendicular
to every straight line in that plane
meetmgit(3. def. 11.); and con-
sequently it is perpendicular, to
CE: wherefore ABF is a right
angle; but GFB is likewise a
right angle : therefore AB is paral-
lel (28. 1.) to FG. And AB is at right angles to the idane CK:
therefore FG is also at right angles to the same plane (8. 11.). But
one plane is at right angles to another plane when the straight lines
drawn in one of the planes at right angles to their, common section,
are also at right angles to the other plane (4. def. 11.): and any
straight line FG in the plane DE, which is at right angles to CE the
common section of the planes, has been proved to be perpendicular
to the other plane CK ; therefore the plane DE is at right angles to
the plane CK. In like manner, it may be proved that all the planes
which pass through AB are at right angles to the plane CK. There-
fore, if a straight line, &c. CI. E. D.

168

THE ELEMKNTB OF EUCLID.

PROP. XIX. THEOR.

BOOK Xi.

If two planes cutting one another be each of them perpendi-
cular to a third plane ; their common section shall be perpendi-
cular to the sdme plane.

Let the two planes AB, BC be each of them perpendicular to a
third plane, and let BD be the common section of the first two ; ED
is perpendicular to the third plane.

If it be not, from the point D draw in the plane AB, the straight
line DE at right angles to AD, the common section of the plane AB
with the third plane ; and in the plane BC draw DF at right angles
to CD the common section of the plane BC with the third plane.
And because the plane AB is perpendicular to B

the third plane, and DEis drawn in the plane
AB at right angles to AD their common sec-
tion, DE is perpendicular to the third plane (4.
def. 11.). In the same manner, it may be
proved that DF is perpendicular to the third
plane. Wherefore, from the point D two
straight lines stand at right angles to the third
plane, upon the same side of it, which is im-
possible (13. 11.) : therefore, from the point D
there cannot be any straight line at right an-
gles to the third plane, except BD the common
section of the planes AB, BC. BD therefore is ^ ^

perpendicular to the third plane. Wherefore, if two planes, &c.
a.E. D.

PROP. XX. THEOR.

If a solid angle be contained by three pkne angles, any two
of them are greater than the third.*

Let the solid angle at A be contained by the three plane angles,
BAC, CAD, DAB. Any two of them are greater than the third.

If the angles BAC, CAD, DAB be all equal, it is evident that '
any two of them are greater than the third. But if they be not,
let BAC be that angle which is not less than either of the other
two, and is greater than one of them DAB ; and at the point A
in the straight line AB, make, in the plane which passes Uiroug^
BA, AC, the angle BAE equal (23. 1.) to the angle DAB; and
make AE equal to AD, and through E draw BEC cutting AB,
AC in the points B, C, and join DB, DC. And because DA is

* See Note.

J*-

BOOK tU

TBB KLNiSNTS OT EVGUD.

equal to AE, and AB is common, the two D

DA, AB are equal to the two £A, AB, and
the angle DAB is equal to the angle EAB :
therefore the base DB Is equal (4. L) to the
base BE. And because BD, DC are great-
er (20. 1.) than CB, and one of them, BD,
has been proved equal to BE a part of CB,
therefore th^ other DC is greater than the ^
remaining part EC. And because DA is
equal to EA, and AC common, but the base DC greater than the base
EC. : therefore the angle DAC is greater («5. 1.) than the angle EAC :
and, by the construction, the angle DAB is equal to the angle BAE ;
wherefore the angles DAB, DAC are together greater than BAE,
EAC, that is, than the angle BAC. But BAC is not less than either
of the angles DAB, BAC : therefore BAC, with either of tiiem, is
greater than the other. Wherefore, if a solid angle, &e» €1. E. D.

PROP. XXL THEOR.

Evert solid angle is contained by plane angles which together
are less than four right angles.

First, let the solid angle at A be contained by three plane angles
BAC, CAD, DAB. These three together are less than four right an-
gles.

Take in each of the straight lines AB, AC, AD any points B, C, D,
and join BC, CD, DB : then, because the solid angle at B, is contain-
ed by the three plane angles CBA, ABD, DBC, any two of them are
greater (20. 11.) than the third; therefore the angles CBA, ABD, are
greater than the angle DBC ; for the same reason, the angles BCA,
ACD are greater than the angle DCB; and the angles CDA, ADB«
^eater than BDC : wherefore the six angles CBA, ABD, BCA, ACD,
CDA, ADB, are greater than the three an- D

gles DBC, BCD. CDB ; but the three angles
DBC, BCD, CUB, are equal to two right an-
gles (32. 1.): therefore the six angles CBA,
ABD, BCA, ACD, CDA, ADB are greater
than two right angles : and because the
three angles of each of the triangles ABC,
ACD, ADB are equal to two right angles,
therefore the nine angles of these three triangles, viz. the angles
CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to
six right angles ; of these the six angles CBA, ACB, ACD, CDA,
ADB, DBA are greater than two right angles : therefore the remain-
ing three angles BAC, DAC, BAD, which contain the solid angle at
A, are less than four right angles.

Next, let the solid angle at A be contained by any number of plane
angles BAC, CAD, DAE, EAP, FAB ; these together are less than
lour right angles.

Let the planes in which the angles are, be cut by a plane, and let

170 THB ELEMGNT8 OF SVCUO. 900K XI.

■*

the common section of it with those planes be BC, CD, DB, EB!,
FB ; and because the solid angle at B is contained by three plane
angles CBA, ABF, FBC, of which any two are greater (20- 1 1.) than
the third, the angles CBA, ABF, are greater than the angle FBC : for
the same reason, the two plane angles at
each of the points C, D, E, F, viz. the an-
gles which are at the bases of the trian^es,
having the common vertex A, are greater '
than the third angle at the same point,
which is one of the angles of the polygon ^
6CDEF: therefore all the angles at the
bases of the triangles are together greater
than all the angles of the polygon : and be-
cause all the angles of the triangles are to-
gether equal to twice as many right angles
as there are triangles (32. 1.); that is,
as there are sides in the polygon BCDEF :
and that all the angles of the polygon, together with four right an-
gles, are likewise equal to twice as many right angles as there are
sides in the polygon (1. Cor. 32. 1.) : therefore all the angles of the
triangles are equal to all the angles of the polygon together with four
right angles. But ail the angles at the bases of the triangles are
greater than all the angles of the polygon, as has been proved.
Wherefore, the remaining angles of the triangles, viz. those at the
vertex, which contain the solid angle at A, are less than four right
angles. Therefore every solid angle, &c. €t £. D.

PROP. XXn. THEOR.

If every two or three plane angles be greater than the third,
and if the straight lines which contain them be all equal ; a tri«
angle may be made of the straight lines, that join the extremities
of those equal straight lines.*

Let ABC, DEF, GHK be three plane angles, whereof every two
are greater than the third, and are contained by the equal straight
lines AB, BC, DE, EF, GH, HK ; if their extremities be joined by the
straight lines AC, DF, GK, a triangle may be made of three straight
lines equal to AC, DF, GK ; that is, every two of them together great-
er than the third.

If the angles at B, E, H are equal ; AC, DF, GK are also equal
(4. 1.), and any two of them greater than the third : but if the
angles be not all equal, let the angle ABC be not less than either
of the two at E, H ; therefore the straight line AC is not less than
either of the other two DF, GK (4. Cor. 24. 1.); and it is plain
that AC, together with either of the other two, roust be greater
than the third : also, DF, with GK, are greater than AC : for at
the point B in the straight luie AB make (23. 1.) the angle ABL

» See Note.

BOOK ZI4

T8V BLBM nrrs of BIIOLm.

171

equal to the angle QHK, and make BL equal to one of the straight
lines AB, BC, DE, EF, GH, HK, and join AL, LC ; then because
AB, BL are equal to GH, HK, and the angle ABL to the angle GHK,
tiie base AL is equal to the base GK ; and because the angles at E,
H are greater than the angle ABC, of which the angle at H is equal
to ABL ; therefore the remaining angle at E is greater than the an-
gle LBC ; and because the two sides jiB, BC are equal to the two

B £

DE, EF, and that the angle DEF is greater than the angle LBC, the
base DF is greater (24. 1.) than the base LC: and it has been
proved that GK is equal to AL ; therefore DF and GK are greater
than AL and LC; but AL and LC are greater (20. 1.) than AC:
much more then are DF and GK greater than AC. Wherefore
every two of these straight lines AC, DF, GK are greater than the
third ; and, therefore, a triangle may be made (22. 1.) the sides of
which shaD be equal to AC, DF, GK. Q. E. D.

PROP. XXm. PROB.

To make a solid angle which shall be contained by three
given plane angles, an v two of them being greater than the third,
and all three together less than four right angles.*

Let the three given plane angles be ABC, DEF, GHK, any two of
which are greater than the third, and all of them together less than
four right angles. It is required to make a solid angle contained by
three j^lane angles equal to ABC, DEF, GHK, each to each.

B H

« See Note,

THfi Bii£M£HT8 OF BUCI*ID*

BOOK XI.

From the straight lines containing tlie angles, cut off AB, BC, DS,
EF, GH, HK, all equal to one another ; and join AC, DP, GK ; then
a triangle may be made (22. li.) of three straight lines equal to AC^
5f, GK. Let this be the triangle LMN (22. 1.) so that AC be equal
to LM, DF to MN, and GK to LN ; and about the triangle LMN
describe (5. 4.) a circle^ and find its cenlare X, which wiil either be
within the triangle, or in one of its sides, (»* without it.

First; let the centre X be "Within the triangle, and join LX, MX,
NX : AB is greater than LX ; if not, AB must either be equal to,
or less, than LX ; .first let it be equal : then because AB is equal to
LX, and that AB is also equal to BC, and LX to XM, AB and BC
are equal to LX and XM, each to each ; and the base AC is, by
construction, equal to the base LM : wherefore the angle ABC is
equal to the angle LXM (8. 1.)* For the same reason, the angle
DEF is equal to the angle MXN, and R

the angle GHK to the angle NXL ; there-
fcNre the three angles ABC, DEF, GHK
are equal to the three angles LXM,
MXN, NXL : but the three angles LXM,
MXN, NXL are equal to four right an-
gles (2. Cor. 15. 1.): therefore also the
three angles ABC, DEF, GHK, are equal
to four right angles; but by the hypo-
thesis, they are less than four right an-
gles ; which is absurd ;. therefore AB is
not equal to LX : but neither can AB
be less than LX : for, if possible, let it
be less, and upon the straight line LM,

the side of it on which is the centre X, describe the triangle LOM,
the sides LO, OM of which are equal to AB, BC ; and because the
base LM is equal to the base AC, the angle LOM is equal to the
angle ABC (8. 1.) : and AB, that is LO, by the hypothesis, is lessr
than LX ; wherefore LO, OM fan within the triangle LXM ; for if
they fell upon its sides, or without it, they would be equs^ to, or
greater than LX, XM (2L 1.) : therefore the angle LOM, that is the
angle ABC, is greater than the angle LXM (2L L): in the same

manner it may be proved that the
angle DEF is greater than the angle
MXN, and the angle GHK greater
than the angle NXL. Therefore the
three angles ABC, DEF, GHK are
greater than the three angles LXM,
MXN, NXL ; that is, than four right
angles; but the same angles ABC,
DEF, GHK are less than four right
angles ; whidi is absurd : therefore
AB is not less than LX, and it has
been proved that it is not equal to
LX ; wherefore AB is greater than
LX.
Next, let the centre X of the circle

BOOK XE»

TRS U«EltEllT8 OF BUGUIH

IT?

faU in one of the sides of the triangle, viz. in MN, and Join XL: in

this case also AB is greater than LX. If not, AB is either equal to

LX, or less than it; first, let it be equal to XL: therefore AB 9Qa

BC, that is, DE and EF, are equal to R

MX and XL, that is, to MN : but by

the construction, MN is equal to DF ;

therefore DE, EF are equal to DF,

which is Impossible (20. 1.) : wherefore

AB is not equal to LX ; nor is it less ;

for then, much more, an absurdity
would follow : therefore AB is greater
than LX. • M

But let the centre X of the circle fall
without the triangle LMN, and join LX,
MX, NX. In this case likewise AB is
greater than LX: if not, it is either
equal to or less than LX : first, let it be equal ; it may be proved in
the same manner, as m the first case, that the angle ABC is equal
to the angle MXL, and GEEK to LXN ; therefore the whole angle
MXN is equal to the two angles ABC, QHK ; but ABC and GHK
are together greater than the angle DEF ; therefore also the angle
MXN is greater than DEF. And because DE, EF are equal to IVOC,
XN, and the base DF to the base MN, tl^e angle MXN is equal (8. 1.)
to the angle DEF : and it has been proved that it is greater than
DEF, which is absurd. Therefore AB is not equal to LX. Nor yet
is it less ; for then, as has been proved in the first case, the angle
ABC is greater than the angle MXL, end the angle GHK greater
than the angle LXN. At the point B in the straight line CB make

B H

the angle CBP equal to the angle GHK, and make BP equal to HKr
and join CP, AP. And because CB is equal to GH; CB, BP arc
equal to GH, HK, each to each, and they contain equal angles ;
wherefore the base CP is equal to the base GK, that is, to LN. And
in the isosceles triangles ABC, MXL, because the angle ABC is
greater than the angle MXL, therefore the angle MLX at the base is
is greater (32. L) than the angle ACB at the base. For the same
reason, because the angle GHK, or CBP, is greater than the an^e
LXN, the angle XLN, is greater than the angle CBP. Therefore
the whole angle MLN is greater than the whole angle ACP. And
beeause ML> LN are equd to AC> CP, each to each, but the angle

174

THft ELEBifiNTa OF SOCLID.

SOQK Zf»

MLN is greater than the angle ACP, the R

base MN is greater (24. 1.) than the base
AP. And MN is equal to DF ; therefore
also DF is greater than AP. Again,
because DE, EF are equal to AB, BP,
but the base DF greater than the base
AP, the angle DEF is greater (25. i.) tiian
the angle ABP. And ABP is equal to
the two angles ABC, CBP, that is, to thie
two angles ABC, GHK; therefore theM
angle DEF is greater than the two an-
gles ABC, GHK ; but it is also less than
these; which is impossibJe. Therefore
AB is not less than LX, and it has been
proved that it is not equal to it ; therefore
AB is greater than LX.

FVom the point X erect (12. 11.) XR at right angles to the plane
of the circle LMN. And because it has been proved in all the cases,
that AB is greater than LX, find a square equal to the excess of the
square of AB above the square of LX,
and make RX, equal to its side; and join
RL, RM, RN. Because RX is perpen*
dicularto the .plane of the circle LMN,
it is (3. def. 1 L) perpendicular to each of
the straight lines LX, MX, NX. And
because LX is equal to MX, and XR
common, and at right angles to each of
them, the base RL is equal to the base M
RM. For the same reason, RN is equal to
each of the two RL, RM. Therefore
the three straight lines RL, RM, RN are
ell equal. And because the square of
XR is equal to the excess of the square
of AB above the square of LX ; there-
fore the square of AB is equal to the squares of LX, XR. But
the square of RL is equal (47. 1.) to the same squares, because
LXR is a right angle. Therefore the square of AB is equal to the
square of RL, and the straight line AB to RL. But each of the
straight lines BC, DE, EF, GH, HK is equal to AB, and each of the
two RM, RN, is equal to RL. Wherefore AB, BC, DE, EF, GH, HK
are each of them equal to each of the straight lines RL, RM, RN.
And because, RL, RM are equal to AB, BC, and the base LM to the
base AC ; the angle LRM is eqiial (8. 1.) to the angle ABC. For
tbe same reason, the angle MRN is equal to the angle DEF, and
NRL to GHK. Therefore there is made a solid angle at R, which is
contained by three plane angles LRM, MRN, NRL, which are equal
to the three given plane angles ABC, DEF, GHK, each to each.
MTfaich was to be done.

BOOK SI. THB BLBMBMTS OV EUCUD. 175

PROP. A. THEOR.

Ip each of two solid angles be contained by three plane angles
equal to one another, each to each ; the planes in which the equal
angles are, have the same inclination to one another.*

Let there be two solid angles at the points A, B ; and let the angle at
A be contained by the three plane angles CAD, C AE, BAD ; and the
angle at B by the plane angles FBG, FBH, HBG, of which the angle
CAD is equal to the angle FBG ; and CAE to FBH ; and EAD to
HBQ : the planes in which the equal angles are, have the same inclir
nation to one another. '

In the straight line AC take any point K, and in the plane CAD
from K draw the straight line KD at right angles to AC, and in the
plane CAE the straight A B

line KL at right angles
to the same AC : there-
fore the angle DKL is the
inclination (9. de£ 11.) of
the plane CAD to the
plane CAE. In BF take

BM equal to AK, and / \ ^ ri / ' I ^ G

from the point M draw,
in the i^anes FBG, FBH,
the straight lines MG, MN at right angles to BF ; therefore the angle
GMN is the inclination (6. def. 11.) of the plane FBG to the plane
FBH ; join LD, NG ; and because in the triangles KAD, MBG, the
angles KAD, MBG are equal, as also the right anglei^ A^, BMG,
and that the sides AK, BM, adjacent to the equal angles,* are equal to
one another; therefore KD is equal (26. 1.) to MG, and AD to BG:
for the same reason, in the triangles ILA.L, MBN, KL is equal to MN,
and AL to BN: and in the triangles LAD, NBG, LA, AD are equal
to NB, BG, and they contain equal angles ; therefore the base LD is
equal (4. 1.) to the base NG. Lastly, in the triangles KLD, MNG,
the sides DK, KL are equal to GM, MN, and the base LD to the
base NG : therefore the angle DKL is equal (8. 1.) to the angle GMN :
but the angle DKL is the inclination of the plane CAD to the plane
CAE, and the angle GMN is the inclination of the plane FBG to the
plane FBH, which planes have therefore the same inclination (7. def.
1 1.) to one uiother : and in the same manner it may be demonstrated,
that the other planes in which the equal angles are, have the same
inclination to one another. Tharefore, if two solid angles, &c. Q. E. D.

PROP. B. THEOR.
Ir two solid angles be contained, each by three plane angles

« See Note.

176 THE ELEMENTS OP EDCLID. BdOK XI.

which are equal to one another, each to each, and alike situated ;
these solid angles are equal to one another."*^

Let there be two solid angles at A and B, of which the solid angle
at A is contained by the three plane angles CAD ; CAE, BAD ; and
that at B, by the three plane angles FBG, FBH, HBG ; of which CAD
IS equal to FBG ; CAE to FBH ; and EAD to HBG : the solid ai^le
at A is equal to the solid angle at B.

Let the solid angle at A be applied to the solid angle at B ; and,
first, the plane angle CAD being applied to the plane angle FBG, so
as the point A may coincide with the point B, and the straight line
AC with BF ; then AD coincides with BG, because the angle GAD
is equal to the angle FBG ; and because
the inclination of the platie CAE to the
plane CAD is equal (A. 11.) to the incli-
nation of the plane FBH to the plane
F.BG, the plane CAE coincides with the
plane FBH, because the planes CAD, E ^ F

FBG coincide with one another : and be- ^ ^ ^

cause the straight lines AC, BF coincide, and that the angle CAE is
equal to the angle FBH; therefore AE coincides with BH, and AD
coincides with BG; wherefore the pfane EAD coincides with the
plane HBG : therefore the solid angle A coincides with the solid an-
gle B, and consequently they are equal (A. 8. 1.) to one another.
Q. E. D.

^ PROP. C. THEOR.

.« •
Solid figures contained by the same nunnber of equal and simi-
lar planes alike situated, and having none of their solid angles
contained by more than three plane angles ; are equal and similar
to one another.*

Let AG, Ed be two solid figures contained by the same number
of similar and equal planes, alike situated, viz. let the plane AC be
similar and equal to the plane KM ; the plane AF to KP ; BG to LQ;
GD to QJV ; DE to NO ; and lastly, FH similar and equal to PR : the
solid figure AG is equal and similar to the solid figure EGL

Because the solid angle at A is contained by the three plane
angles BAD, BAE, EAD, which, by the hypothesis, are equal to
the plane angles LKN, LEO, OEJV, which contain the dolid angle
at E, each to each ; therefore the solid angle at A is equal (B. 11.)
to the solid angle at E : in the same manner, the other solid angles
of the figures are equal to one another. If, then, the solid figure
AG be applied to the solid figure Ed, first, the plane figure AC

* » See Note.

BOOK XL

THS EUBMfiMTS OF BUCLtD.

177

being applied to the
plane figure EM :
the straight line AB
coinciding with EL
the figure AC must
coincide with the fi-
gure E[M, because
they are equal and

^NI

similar : th^efore the straight lines AD, DCt CB, coincide with EN,
NM, ML, each with each ; and the points A, D, C, B, with the points
E, N, M, L: and the solid angle at A coincides with (B. 11.) the
solid angle at E ; wherefore the plane AF coincides witli the plane
EP, and the figure AF with the figure EP, because they are equal
and similar to one another : therefore the straight lines AE, £F, FB,
coincide with EO, OP, PL ; and the points E, F, with the points O,
P. In the same manner, the figure AH coincides with the figure
ER, and the straight line DH with NR, and the point H with the
point R : and because the solid angle at B is equal to the solid angle
at L, it may be proved, in the same manner, that the figure BG coin-
cides with the figure Ld, and the straight line CG with MQ, and
the point G with the point Q : since, therefore, all the planes and
sides of the solid figure AG coincide with the planes and sides, of the
solid ^gixre EQ, AG is equal and similar to EQ,: and, in the same
manner, any other solid figures whatever contained by the same
number of equal and similar planes, alike situated, and having none
of their solid angles contained by more thaf three plane angles,, may
be proved to be equal and similar to one another. Q. E. D.

PROP. XXIV. THEOR.

If a solid be contained by six planes, two and two of which are
parallel ; the opposite planes are similar and equal parallelograms.*

Let the solid CDGH be contained by the parallel planes AC, GF ;
BG, CE ; FB, AE : its opposite planes are similar and equal paral-
lelograms.

Because the parallel planes BG, CE are cut by the pJane AC, their
common sections AB, CD are parallel (16. 11.). Again, because the
two parallel planes BF, AE are cut by the plane AC, their common
section AD, BC are parallel (16. 11.}; and AB is parallel to CD;
therefc»*e AC is a parallelogram. In Hke manner it may be proved

that each of the figures CE, FG, GB, BF,
AE is a paralldlelogram : join AH, DF ;
and because AB is parallel to DC, and
BH to CF ; the two straight lines AB,
BH, which meet one another, are paral-
lel to DC and CF which meet one an-
other, and are not in the same plane
with the other two : wherefore they con-
tain equal angles (10. 11.); the angle

* See Note.
23

=5F

176

THB KLEMBNI'S OF EUCLID.

BOOK XI.

ABH is therefore equal to the angle DCF j and because AB, BH are
equal to DC, CF, and the angle ABH equal to the angle DCF ; there-
fore the base AH is equal (4. 1.) to the base DF, and the triangle ABH
to the triangle DCF : and the parallelogram BG is double (34. I.)
of the triangle ABH, and the parallelogram CE double of the triangle
DCF ; therefore the parallelogram BG is equal and similar to the
parallelogram CE. In the same manner it may be proved, that the
parallelogram AC is equal and similar to the parallelogram OF, and
the parallelogram AE to BF. Therefore, if a solid, &c. Q. E, D.

PROP.^XXV. THEOR.

If a solid parallelepiped be cut by a plane parallel to two of
its opposite planes, it divides the whole into two solids, the base
of one of which shall be to the base of the other, as the one solid
is to the other.*

Let the solid parallelopiped ABCD be cut by the plane EV, which
is parallel to the opposite planes AR, HD ; and divides the whole
into the two solids ABFV, EGCD ; as the base AEFY of the first is
to the base EHCF of the other ; so is the solid ABFV to the solid
EGCD. .

Produce AH both ways, and take any number of straight lines
HM, MN, each equal to EH, and any number AK, KL each equal
to EA, and complete the parallelograms LO, KY, HQ, MS, and the
solids LP, KR, HU, MT* then because the straight . lines LK, KA,
AE are all equal, the parallelograms LO, KY, AF are equal (36. 1.) :

O Y

c a

and likewise the parallelograms KX, BK, AG, (36. L) ; as also (24.
11.) the parallelograms LZ, 'KP, AR, because they are opposite
plai)es : for the same reason the parallelograms EC, HQ, MS are equad
(36. 1 .) ; and the paraUelograms HG, HI, IN, as also <24. 1 1 .) HD, MU,
NT ; therefore three planes of the solid LP, are equal and similar to
three planes of the solid KR, as also to three planes of the solid AV :
but the three planes opposite to these three are equal and similar (24,
11.) to them in the several solids, and none of their solid angles are
ccmtained by more than three plane angles : therefore the three sa&da
LP, KB, AV are equal (C. 11.) to one another: for the same reason,
the three solids ED, HU, MT are equal to one another : therefcara

» See Note.

BOOK XI.

THE ELEMENTS OF EUCLID.

179

what multiple soever the hase LF is of the base AF, the same multi-
ple is the solid LV of the solid AV : for the same reason, whatever
multiple the base NF is of the base HF, the same multiple is the
solid NV of the solid ED ; and if the base LF, be equal to the
base NF, the solid LV is equal (C. IL) to the sdid NV; and if
the base LF be greater than the base NF, the solid LV is greater
than the solid NV; and if less, less: since then there are four
magnitudes, viz. the two bases AF, FH, and the two solids AV,

X B G I

KHTM

O Y F c a s

ED, and of the base AF and solid AV, the base LF and solid LV
are any equimultiples whatever ; and of the base FH and solid ED,
the base FN and solid NV are any equimultiples whatever ; and it
has been proved, that if the base LF is greater than the base FN,.
the solid LV is greater than the solid NV ; and if equal, equal; and
if less, less. Therefore (5. def. 5.) as the base AF is to the base
FH, so is the solid AV to the solid ED. Wherefore, if a solid, &c.
a. E. D.

PROP. XXVI. PROB.

At a given point in a given straight line, to make a solid an-
gle equal to a given solid angle contained by three plane angles-*

Let AB be a given straight line, A a given point in it, and D a
given solid angle contained by the three plane angles EDC, EDF,
FDC : it is required to make at the point A in the straight line AB
a solid angle equal to the solid angle D.

In the straight line DF take any point F, from which draw (IL^
IL) GF perpendicular to the plane EDC, meeting ^hat plane in G;
join DQ, and at the point A in the straight line AB make (23. L)
the angle BAL equal to the angle EDC, and in the plane BAL make
the angle BAK equal to the angle EDG : then make AK equal to
DG, and from the poiftt K erect (12. U.) KH at right angles to the
plane BAL; and make KH equal to GF, and join AH: then the
solid angle at A, which is contained by the three plane angles BAL,
BAH, HAL, is equal to the solid angle at D contained by the three
plane angles EDC, EDF, FDC.

Take the equal straight lines AB, DE, and join HB, KB, FE,
GE: and because FG is perpendicular to the plane EDC, it
makes right angles (3. def. 11.) with every straight line meeting

* See Note.

180 THE ELEMEMTB OF EUCLID. BOOK XI.

It in that plane : therefore each of the angles FGD, FGE is a right
angle : for the same reason, HKA, HKB, are right angles : and be-
cause KA, AB are equal to GD, DE, each to each, and contain equal
angles, therefore the base BE is equal (4. I.) to the base EG: and
KH is equal to GF, and HEB, FGE are right angles, therefore HB is
equal (4. 1.) to FE: again, because AK, KH are equal to DG, GF,
and contain right angles, the base AH is equal to the base DF ; and
AB is equal to DE ; therefore HA, AB are equal to FD, DE, and the
base HB is equal to the
base FE, therefore the an-
gle BAH is equal (8.' 1.)
to the angle EDF : for the
same reason, the angle
HAL is equal to the an-
gle FDC. Because if AL

and DC be made equal, k '<^S^ ^ ^ q

and KL, HL, GC, FC be H F

joined, since the whole angle GAL is equal to the whole EDO, and
the parts of them BAE, EDG are, by the construction, equal: there-
fore the remaining angle EAL is equal to the remaining angle GDC :
and because EA, AL are equal to GD, DC, and contain equal angles,
the base EL is equal (4. L) to the base GC : and EH is equal to GF,
so that LE, EH are equal to CG, GF, and they contain right angles :
therefore the base HL is equal to the base FC : again, because HA,
AL are equal to FD, DC, and the base HL to the base FC, the angle
HAL is equal (8. 1.) to the angle FDC : therefore, because the three
plane angles BAL, BAH, HAL, which contain the solid angle at A,
are equal to the three plane angles EDC, EDF, FDC, which contain
the solid angle at D, each to each, and are situated in the same order,
the solid angle at A is equal (B. 11.) to the solid angle at D. There-
fore, at a given point in a given straight line, a solid angle has been
made equal to a given solid angle contained by three plane angles.
Which was to be done.

PROP. XXVn. PROB.

To describe from a given straight line a solid parallelepiped
similar and similarly situated to one given.

Let AB be the given straight line, and CD the given solid parallelo-
piped. It is reqnured from AB to describe a solid parallelopiped si-
milar and similarly situated to CD.

At the point A of the given straight line AB, make (26. IL)
a solid angle equal to the solid angle at C ; and let BAE, EAH,
HAB, be the three plane angles which contain it, so that BAE
be equal to the angle ECG, and EAH to GCF, and HAB to
FCE: and as EQ to CG, so make (12. 6.) BA to AE: and as
GC to CF, so make (12. 6.) EA to AH ; wherefore ex xquali (22.
<5.) as EC to CF, so is BA to AH; complete the parsdlelogram

BOOK Xt.

THE ELEMENTS OF EUCLID.

181

BH, and the solid AL:
and because, as EC to
CG, so BA to.AK, the
sides about the equal an-
gles ECG, BAK are pro-
portionals : therefore the
parallelogram BK is simi-
lar to EG. For the same
reason, the parallelogram
EH is similar to GF, and

HB to FK Wherefore three parallelograms of the solid AL are
similar to three of the solid CD ; and the three opposite ones in each
solid are equal (24. 11.) and similar to these, each to each. Also,
because the plane angles which contain the solid angles of the figures
are equal, each to each, and situated in the same order, the solid an-
gles are equal (B. 1 1.) each to each. Therefore the solid AL is simi-
lar (11. def. 11.) to the solid CD. Wherefore from a given straight
line AB, a solid parallelepiped AL has been described similar and
similarly situated to the given one CD. Which was to be done.

PROP. XXVin. THEOR.

If a solid parallelopiped be cut by a plane pssing through the
diagonals of two of the opposite planes ; it snail be cut into two
equal parts.^

Let AB be a solid parallelopiped, and DE, CF the diagonals of the
opposite parallelograms AH, GB, viz. those which are drawn betwixt
the equal angles in each : and because CD, FE are each of them pa-
rallel to GA, and not in the same plane with it, CD, FE are parallel
(9. 11.) ; wherefore the diagonals CF, DE are in the plane in which
the parsdlels are, and are themselves paral- C B

lels (16. 11.); and the plane CDEF shall cut
the solid AB into two equal parts.

Because the triangle CGF is equal (34. 1.)
to the triangle CBF, and the triangle DAE,
to DHE ; and that the parallelogram CA is
equal (34. 11.) and similar to the opposite
one BE ; and the parallelogram GE to CH :
therefore the prism contained by the two
triangles CGF, DAE, and the three parallel-
ograms CA, GE, EC, is equal (C. 11.) to the prism contained by the
two triangles CBF, DHE, and the three parallelograms BE, CH, EC ;
because they are contained by the same number of equal and similar
planes, alike situated, and none of their solid angles are contained by
more than three plane angles. Therefore the solid AB is cut into
two eqval parts by the plane CDEF. €1. E. D.

< N. B. The insisting straight lines of a parallelopiped, mentioned

•See Note.

182

THE BLEM£1IT3 OP EUCLID.

BOOK XI.

in the next and sooie following propositions, are the sides of the pa-
rallelograms betwixt the base and the opposite plane parallel to it*

PROP. XXIX. THEOR.

Solid parallelepipeds upon the same base, and of the same
altitude, the insisting lines of which are terminated in the same
straight lines in the plane opposite to the base, are equal to one
another.*

Let the solid parallelepipeds AH, AK be upon the same base AB,
and of the same altitude, and let their insisting straight lines AF,
AG, LM, LN be terminated in the same straight line FN, and CD,
CE, BH, BK be terminated in the same straight line DK ; the solid
AH is equal to the solid AK.t

First, let the parallelograms DG, HN, which are opposite to the
base AB, have a common side HG : then, because the sdid AH is
cut by the plane AGHC passing through the diagonals AG, CH of
the opposite planes ALGF, CBHD, AH is cut into two equal parts
(28. 11.) by the plane AGHC : therefore the solid AH is double of the
prism whiqfi is contained betwixt the triangles ALG, CBH; for the
same reason because the solid AE D H K

is cut by the plane LGHB through
the diagonals LG, BH of the oppo-
site planes ALNG, CBKH, the solid
AK is double of the same prism
which is contained betwixt the tri- C
angles ALG, CBH. Therefore the
solid AH is equal to the solid AK. ^

But let the parallelograms DM, EN opposite to the base have
no common side : then, because CH, CK are parallelograms, CB is
equal (34. 1.) to each of the opposite sides DH, EK; wherefore
DH is equal to EK: add or take away the common part HE; then
DE is equal to HK : wherefore also the triangle CDE is equal
(38. 1.) to the triangle BHK: and the parallelogram DC is equal
(36. 1.) to the parallelogram HN : for the same reason the triangle
D HE KDE HK

A L A L

AFG is equal to the triangle LMN, and the parallelogram CF
is equal (24. 11.) to the paralldogram BN, and CG to BN; for
they aire opposite. Therefore the prism which is contained by

« See Note.

t See the figures.

BOOK XI. THB ELEMENTS OF EUCLID. 1S3

the two triangles AFG, ODE, and the three parallelograms AD, DO,
GO, is equal (C. 11. to the prism contained by the two triangles
LMN, BHK, and the three parallelograms BM, MK, KL. If therefore
the prism LMNBHE be taken from the solid of which the base is
the parallelogram AB, and in which FDEN is the one opposite to it;
and if from this same solid there be taken the prism AFGCDE, the
remaining solid, viz. the parallelopiped AH, is equal to the remaining
paraUelopiped AE. Therefore, solid parallelopipeds, &c. Q. E. D.

PReP. XXX. THEOR.

Solid parallelopipeds upon the same base, and of the same
altitude, the insisting straight lines of which are not. terminated in
the same straight lines in the plane opposite to the base, are equal
to one another.*

Let the parallelopipeds CM, CN be upon the same base AB, and of
the same altitude, but their insisting straight lines AF, AG, LM, LN,
CD, CE, BH, BK, not terminated in the same straight lines ; the
solids CM, CN, are equal to one another.

Produce FD, MH, and NG, KE ; and let them meet one another in
the points O, P, ^ R ; and join AO, LP, BQ, CR : and because the
plane LBHN is parallel to the opposite plane ACDF, and that the

A C

plane LBHM is that in wjiich are the parallels LB, MHPQ, in which
also is the figure BLPQ,, and the plane ACDF is that in which are
the parallels AC, FDOR, in which also is the figure CAOR ; therefore
the figures BLPQ, CAOR are in parallel planes ; in like manner, be-
cause the plane ALNG is parallel to the opposite plane CBKE, and
that the jdane ALNG is that in which are thie parallels AL, OPGN, in
which also is the figure ALPO ; and the plane CBKE is that in which
are the paraDels CB, RQ^EE, in which also is the figure CBQR;
therefore the figures ALPO, CBQR are in parallel planes ; and the
planes ACBL, ORQP are parallel ; therefore the solid CP is a paral-

* See Note.

184 THB ELEMENTS OF EUCLID. BOOK XL

]ek>piped ; but the solid CM, of which the base is ACBL, to which
FDHM is the opposite parallelogram, is equal (29. 11.) to the solid
CP, of which the base is the parallelogram ACBL, to which ORQP

N K

is the one opposite ; because they are upon the same base, and their
insisting straight lines AF, AO, CD, CR ; LM, LP, BH, BQt are in
the same straight lines FR, MQ,: and the solid CP is equal (29. 11.)
to the solid CN : for they are upon the same base ACBL, and their
insisting straight lines AO, AG, LP, LN ; CR, CE, BQ, BK, are in
the same straight lines ON, RK : therefore the solid CM is equal to
the solid CN. Wherefore, solid parallelepipeds, &c. Q,. E. D.

PROP. XXXL THEOR.

Solid parallelopipeds which are upon equal bases and of the
same altitude, are equal to one another.*

Let the solid parallelopipeds AE, CF be upon equal bases AB, CD,
and be of the same altitude ; the solid AE is equal to the solid CF.

First, let the insisting straight lines be at right angles to the bases
AB, CD, and let the bases be placed in the same plane, and so as that
the sides CL, LB be in a straight line ; therefore the straight luie LM
which is at right angles to the plane in which the bases are, in the
pouit L, is common (13. 11.) tb the two solids AE, CF; let the other
insisting lines of the solids be AG, HK, BE; DF, OP, CN : and first,
let the angle ALB be equal to the angle CLD ; then AL, LD are in
a straight line (14. 1.). Produce OD, HB, and let them meet in Q,
and complete the solid parallelepiped LR, the base of which is the pa*
ralelogram LQ, and of which LM is one of its insisting straight lines:
therefore, because the parallelogram AB is equal to CD ; as the base
AB is to the base Ld, so is (7. 5.) the base CD to the same LQ : and
because the solid paralldiopiped AR is cut by the plane LME^B, vrhkh

* See Note.

BOOK XI.

TIUB ELEMENTS OF EUGUD.

186

is parallel to the opposite ]^anes AK, DR ; as the base AB is to the
base LQ, so is (25. 11.) the solid A£ to the solid LR : for the same
reason, because the solid parallelopiped CR is cut by the plane
LMFD, which is parallel to the opposite planes CP, BR ; as the base

CD to the base LQ, so
is the solid OF to the
solid LR: but as the
base AB to the base
LQ^ so the base CD to
to the base LQ, as be-
fore was proved: there-
fore as the solid AE to
the solid LR, so is the
solid CF to the solid

C L

AS H T

LR; and therefore the solid AE is equal (9. 5.) to the solid CF.

But let the solid parallelopipeds SE, CF be upon equal bases SB,
CD, and be of the same altitude, and let their insisting straight lines
be at right angles to the bases ; and place the bases SB, CD in the
same plane, so that CL, LB be in a straight line ; and let the angles
SLB, CLD be unequal ; the solid SE is also in this case equal to the
solid CF : produce PL, TS, until they meet in A, and from B draw
BH parallel to DA ; and let HB, OD produced meet in Q, and com-
plete the solids AE, LR ; therefore the solid AE, of which the base
is the parallelogram LE, and AK the one opposite to it, is equal (29.
11.) to the solid SE, of which the base is LE, and to which SX is
opposite : for they are upon the same base LE, and of the same alti-
tude, and their insisting straight lines, viz. LA, LS, BH, BT ; MG,
MV, EK, EX, are in the same straight lines AT, GX ; and because
the parallelogram AB is equal (35. 1.) to SB, for they are upon the
same base LB, and between the same parallels LB, AT ; and that
the base SB is equal to the base CD ; therefore the base AB is equal

to the base CD, and 1
the angle ALB is equal
to the angle CLD;
therefore, by the first
case, the solid AE is
equaJ to the solid CF ; O
but the solid AE is
equal to the solid SE2,
as was demonstrated ;
therefore the solid SE
is equal to the solid CF.

\**

"X,

^"^^^^v^

A S

H T

But if the msisting straight Imes AG, HE, BE, LM ; CN, RS, DF»
OP, be not at right angles to the bases AB, CD ; in this case like-
wise the solid AE, is equal to the solid CF : from the points G, K,
E. M ; N, S, F, P, draw the straight lines GQ, KT, EV, MX ; NY,
SZ, FI, PU, perpendicular (11. 11.) to the plane in which are the
bases AB, CD ; and let them meet it in the points Q, T, V, X ; Y, Z,
I, U and join QT, TV, VX, Xa ; YZ, ZI, lU, UY : then because
G€l, KT are at right angles to the same plane, they are parallel (6.

196

THfi ELEMENTS OP EVCLID.

BOOK XI.

AHaT CR Y Z

11.) to one another: and MG, EK are parallels ; therefore the plane
MQ, ET, of which one passes through MG, GQ, and the other
through EK, KT, which are parallel to MG, GQ,, and not in the same
plane with them, are parallel (15. 11.) to one another. For the same
reason the planes M V, GT are parallel to one another : therefore the
solid Q.E is a parallelopiped : in like manner it may be proved, that
the solid YF is a parallelopiped : but, from what has been demon-
strated, the solid EQ, is equal to the solid FY, because they are upon
equal bases MK, PS, and of the same altitude, and have their insist-
ing straight lines at right angles to the bases : and the solid Ed is
equal (29. or 30. 11.) to the solid AE;.and the solid FY to the solid
CF ; because they are upon the same bases and of the same alti-
tude : therefore the solid AE is equal to the solid CF. Wherefore
solid paralleloplpeds, &c. Q. E. D.

PROP. XXXn. THEOR.

Solid parallelopipeds which have the same altitude, are to one
another as their bases.* I

Let AB, CD be solid parallelopipeds of the samealtidude; they are
to one another as their bases ; that is, as the base AE to the base
CF, so is the solid AB to the solid CD.

To the straight line FG apply the parallelogram FH equal (Cor.
45. 1.) to AE, so that the angle FGH be equal to the angle LCG,
and complete the solid parallelopiped GK upon the base FH, one of
whose insisting lines is FD, whereby the solids CD, GK must be of
the same altitude ; therefore the solid AB is equal (31. 11.) to the

solid GK, because
they are upon equal
bases AE, FH, and
are of the same alti-
tude: and because
the solid parallelo-
piped CK is cut
by the plane DG
which is parallel to AM C G H

its opposite planes, the base HF is (25. 11.) to the base FC, as the
solid HD to the solid DC : but the base HF is equal to the base AE,
afid the solid GK to the solid AB : therefore, as the base AE to the
base CF, so is the solid AB to the solid CD. Wherefore solid paral-
leloplpeds, &c. Q. E. D.

* See Nolo.

BOOK XL

THB ausMBirrs of bugud.

187

Cor. From this it is manifest that prisms upon tdangolar bases,
of the same altitude, are to one another as their bases.

Let the prisms the bases of which are the triangles AEM, CFG*
and NBO, PDQ the triangles opposite to them, h&ve the same alti-
tude ; and complete the parallelograms AE, CF, and the solid paral-
lelepipeds A5, CD, in the jQrst of which let MO, and in the other let
GQ be one of the insisting lines. And beeause the solid paral-
lelopipeds AB, CD have the s^me altitude, they are to one another as
the base AE is to the base CF ; wherefore the prisms, which are
their halves (28. 11.) are to one another as the base AE to the base
CF ; that is, as the triangle AEM to the triangle CFG.

PROP. XXXni. THEOR.

Similar solid parallelopipeds are one to another in the triplicate
ratio of their homologous sides.

Let AB, CD be similar solid parallelopipeds, and the side AE ho-
mologous to the side CF : the solid AB has to the solid CD the tri-
plicate ratio of that which AE has to CF.

Produce AE, GE, HE, and in these produced take EK equal to CF,
EL equal to FN, and EM equal to FR ; and complete the parallelo-
gram KL, and the solid KO : because KE, EL are equal to CF, FN,
and the angle KEL, equal to the angle CFN, because it is equa]
to the angle AEG, which is equal to CFN, by reason that the
solids AB, CD are similar ; therefore the parallelogram EL is simi-
lar and equal to the parallelogram CN: for the same reason, the
parallelogram MK is similar and equal to CR; and also OE to FD^
Therefore three parallelograms of the solid EO are equal and similar
to three parallelograms 6 X

of the solid CD ; and
the three opposite ones
in each solid are equal
(24. 11.) and similar to
these : therefore the so-
lid EO is equal (C. 11.)
and similar to the solid
CD: complete the pa-
rallelogram GE and
complete the solids £1X,
LP upon the bases GK,
EL, so that £H be an
insisting straight line in each of them, whereby they must be of the
same altitude with the solid AB : and because the solids AB, CD arQ
similar, and, by permutation, as AE is to CF, so is EG to FN, and so
is EH to FR; and FC is equal to EE, and FN to BL, an4
PR to EM ; therefore as AE to EE, so is EG to EL, and so is HB
to EM : but, as AE to EE, so (1. 6.) is the parallelogram AG to the
parallelogram GE ; and as GE to EL, so is (1. 6.) GE to EL, and as
HE to EM, so (1. 6.) is PE to EM : therefore as the parallelogram
AG to the parallelogram GE, so is GE to EL, and PE to EM : but
as AG to GE, so (25. 11.) is the solid AB to the solid EX; and as
GK to KL, so (26. 11.) is the soUd EX to the soUd PL ; and as PE

188 TBB KLJSMBHTB OF BOCLID. BOOK XI.

to KM, 80 (25. 11.) is the solid PL to the solid KO : and therefore as
the solid AB to the solid EX, so is EX to PL, and PL to EO : but if
four magnitudes be continual proportionals, the first is said to have
to the fourth, the triplicate ratio of that which it has to the second :
therefore the solid AB has to the solid KO the triplicate ratio of that
which AB has to EX : but as AB is to EX, so is the parallelogram
AG to the parallelogram GK, and the straight line AE to the straight
line EK. Wherefore the solid AB has to the solid KO the triplicate
ratio of that which AE has to EK. And the solid KO is equal to the
solid CD, and the straight line EK is equal to the straight line CF.
Therefore the solid AB has to the solid CD the triplicate ratio of that
which the side AE has to the homologous side CF, &c. Q,. E. D.

Cor. From this it is manifest, that, if four straight lines be con-
tinual proportionals, as the first is to the fourth, so is the solid paral-
lelopiped described from the first to the similar solid similarly describ-
ed from the second ; because the first straight line has to the fourth
the triplicate ratio of that which it has to the second.

PROP. D. THEOR.

. Solid parallelopipeds contained by parallelograms equiangular
to one another, each to each, that is, of which the solid angles
are equal, each to each, have to one another the ratio which is
the same with the ratio compounded of the ratios of their sides.^

Let AB, CD be solid parallelopipeds, of which AB is contained by
the parallelograms AE, AF, AG, equiangular, each to each, to the
parallelograms CH, CK, CL, which contains the solid CD. The ratio
which the solid AB has to the solid CD, is the same with that which
Is compounded of the i^tios of the sides AM to DL^ AN to DK, and
AO to DH.

Produce MA, NA, OA, to P, Q,, R, so that AP be equal to DL, AQ
to DK, and AR to DH; and complete the solid parallelepiped AX
contained by the parallelograms AS, AF, AV, similar and equal to
CH, CK, CL, each to each. Therefore the solid AX is equal (C. 11.)
to the solid CD- Complete likewise the solid AY, the base of which
is AS, and of which AO is one of its insisting straight lines. Take
any straight line a, and as MA to AP, so make a to b, and as NA to
AQ, so make b to c ; and as AO to All, so c to d : then because the
parallelogram AE is equiangular to AS, AE is to AS, as the straight
line a to c, as is demonstrated in the 23d prop, book 6 : and the
solids AB, AY, being betwixt the parallel planes BOY, EAS, are of
the same altitude. Therefore the solid AB is to the solid AY, as
(32. 11.) the base AE to the base AS; that is as the straight
line a is to c. And the solid AY, is to the solid AX, as (26.
11.) the base OQ is to the base QR; that is, as the straight

* See Note.

BOOK XL

THB EtBMfeXITS OF EaCLID.

189

t^""

, \

a
b
c
d

C M

~ y

V M ■■

V X

line OA to AR ; that is, as the straight Hne c to the straight line d.
And because the solid AB is to the solid AY, as a is to c, and the
solid AY to the solid AX as c is to d ; ex sequalh the solid AB is to
the solid AX, or CD which is equal to it, as the straight line a is to
d. But the ratio of a to d is said to be compounded (def. A. 5.) of
the ratios of a to b, b to c, and c to d, which are the same with ratios
of the sides MA, to AP, NA to AQ, and OA to AR, each to each.
And the sides AP, AQ,, AR are equal to the sides DL, DK, DH, each
to each. Therefore the solid AB has to the solid CD the ratio which
IS the same with tliat which is compounded of the ratios of the sides
AM to DL, AN to DK, and AO to DH. Q. E. D.

PROP. XXXIV. THEOR.

The bases and altitudes of equal solid parallelopipeds, are re-
ciprocally proportional : and if the bases and altitudes be recipro-
cally proportional, the solid parallelopipeds are equal.*

Let AB, CD be equal solid parallelopipeds ; their bases are recipro-
cally proportional to their altitudes ; that is, as the base EH is to the
base NP, so is the altitude of the solid CD to the altitude of the solid
AB.

First, let the insisting straight lines, AG, EF, LB, HK ; CM, NX,
OD, PR be at right angles to the bases. As the base EH to the base

NP, so is CM to AG. If the K
base EH be equal to the base
NP, then because the solid AB
is likewise equal to the solid
CD, CM shall be equal to AG.
Because if the bases EH, NP be
equal, but the altitudes AG, CM H
be not equal, neither shall the
solid AB be equal to the solid
CD. But the solids are equal, by the hypothesis. Therefore the al-
titude CM is not unequal to the altitude AG ; that is, they are equal.
Wherefore, as the base EH to the base NP, so is CM to AG.

* See Note.

190

THB KLBMBNTS OF BIIVLIP.

BOOK XI.

Next, let the bases EH, NP not be equal ; but EH greater than the
other : since then the solid AB is equal to the solid CD, CM, is there-

fore greater than AG : for if it
be not, neither also, in this
case, would the solids AB,
CD be equal, which, by the
hypothesis, are equal. Make
then CT equal to AG, and
complete the solid parallelo-
piped CV of which the base
is NP, and the altitude CT. H
Because the solid AB is equal
to the solid CD, therefore the

£.

G
\

•K- ■■!

V 1

A E C

solid AB is to the solid CV, as (7. 5.) the solid CD to the solid CV.
But as the solid AB to the solid CV, so (32. 11.) is the base EH to
the base NP; for the solids AB, CV are of the same altitude; and as
the soHd CD to CV, so (25. 11.) is the base MP to the base PT, and
so (1. 6.) is the straight line MC to CT ; and CT is equal to AG.
Therefore, as the base EH to the base NP, so is MC to AG. Where-
fore, the bases of the solid parallelepipeds AB, CD are reciprocally
proportional to their altitudes.

Let now the bases of the solid parallelopipeds AB, CD be recipro-
cally proportional to their altitudes ; viz. as the base EH to the base

^T"

NP, so the altitude of the solid
CD to the altitude of the solid
AB ; the solid AB is equal to the
solid CD. Let the insistmg
lines be, as before, at right an-
gles to the bases. Then, if the
base EH be equal to the base H
NP, since EH is to NP, as the
altitude of the solid CD is to the
altitude of the solid AB, therefore the altitude of CD is equal (A. 5.)
to the altitude of AB. But solid parallelopipeds upon equal bases,
and of the same altitude, are equal (31. 11.) to one another : therefore
the solid AB Is equal to the solid CD.

But let the bases EH, NP be unequal, and let EH be the greater
of the two. Therefore, suice as the base EH to the base NP, so

is CM the altitude of the
solid CD to AG the ald-
tude of AB, CM is great-
er (A. 5.) than AG. Again,
take CT equal to AG, and
complete, as before, the
solid CV, And because
the base EH is to the base
NP, as CM to AG, and
that AG is equal to CTt
therefore the base EH is
to the base NP, as MC to

[SZK

BOOK. XL

THS ELBMBHTft OP EUCLID.

CT. But ae the base EH is to NP» so (d2i IL) is tbe solid AB to
the solid C V ; for the solids AB, C V are of the same altitude ; and
as MC to CT, so is the base MP to the base FT, and the solid QD to
the solid (25. 11.) CV : and therefore as the solid AB to the solid
CV, so is the solid CD to the solid CV ; that is, each of the solids
AB, CD has the same ratio to the solid CV ; and theref(»e the solid
AB is equal to the solid CD.

Second general case. Let the msisting straight lines FE, BL, GA,
KH; XN, DO, MC, RP not be at right angles to the bases of the
solids ; and from the points F, B, K, G ; X, D, R, M draw perpen-
diculars to the planes in which are the bases EH, NP, meeting those
planes in the points S, Y, V, T ; Q, I, ^, Z ; and complete the solids
FV, XU, which are parallelopipeds, as was proved in the last part of
Prop. 31. of this Book. In this case likewise, if the solids AB, CD
be equal, their bases are reciprocally proportional to their altitudes,
viz. the base EH to the base NP, as the altitude of the solid CD to
the altitude of the solid AB. Because the solid AB is equal to the
solid CD, and that the solid BT is equal (29. or 30. 11.) to the solid
BA, for they are upon the same base FE, and of the same altitude;
and that the solid DC is equal (29. or 30. 11.) to the solid DZ, being

upon the same base XR, and of the same altitude ; therefore the
solid BT is equal to the solid DZ : but the bases are reciprocally
IM'oportional to the altitudes of equal solid parallelopipeds of which
the insisting straight lines are at right angles to their bases, as be-
fore was proved. ,\ Therefore as the base FK to the base XR, so is
the altitude of the solid DZ to the altitude of the solid BT : and the
base FK is equal to the base EH, and the base XR to the base NP.
Wherefore, as the base EH to the base NP, so is the altitude of the
solid DZ to tbe altitude of the solid BT : but the altitudes of the
solids DZ, DC, as also of the solids BT, BA are the same. There-
fore as the base EH to the base NP, so is the altitude of the solid
CD to the altitude of the solid AB ; that is, the bases of the solid pa-
rallelopipeds AB, CD are reciprocally proportional to their altitud^.
Next, let the bases of the solids AB, CD be reciprocally propor-
tional to their altitudes, viz. the base EH to the base NP, as the alti-
tude of the solid CD to the altitude of the solid AB ; the solid AB is

193

THB ELBME19T8 OF EUGUD.

BOOK XI.

equal to the solid CD : the same construction being made : because
as the base KH to the base NP, so is the altitude of the solid CD to
the altitude of the solid AB; and that the base EH is equal to the
base FK ; and NP to XR ; therefore the base FK is to the base XR,
as the altitude of the solid CD to the altitude of AB. But the alti-
tudes of the solids AB, BT are the same, as also of CD and DZ;

therefore as the base FE to the base XR, so is the altitude of the
solid DZ to the altitude of the solid BT : wherefore the bases of the
solids BT, DZ are reciprocally proportional to their altitudes ; and
their insisting straight lines are at right 'angles to the bases ; where-
fore, as was before proved, the solid BT is equal to the solid DZ :
but BT is equal (29. or 30. 11.) to the solid BA, and DZ to the solid
DC, because they are upon the same bases, and of the same altitude.
Therefore the solid AB is equal to the solid CD. Q,. E. D.

PROP. XXXV. THEOR.

If, from the vertices of two equal plane angles, there be drawn
two straight lines elevated above the planes in which the angles
are, and containing equal angles with the sides of those angles,
each to each ; and if in the lines above the planes there be taken
any points, and from them perpendiculars be drawn to the planes
in which the first named angles are : And from the points in
which they meet the planes, straight lines be drawn to the ver-
tices of the angles first named ; these straight lines shall contain
equal angles with the straight lines which are above the planes
of the angles.*

Let BAC, EDF be two equal plane angles ; and from the points
A, D let the straight lines AG, DM be elevated above the planes
of the angles, making equal angles with their sides, each to each,
viz. the angle GAB equal to the angle MDE, and GAC to MDF :
and in AG, DM let any points G, M be taken, and from them let

* See Note.

BOOK XI.

TBI ELEMENTS OF EOCUD.

193

perpendiculars GL, MN be drawn to the planes BAC, EDP, meettnff
these pilanes in the points L, N, and join LA, ND : the angle GAL i»
equal to the angle MDN.

Make AH equal to DM, and through H draw HK parallel to OL.
But GL IB perpendicular to the plane BAC ; wherefore HK is per-
pendicular (8. 11.) to the same plane : from the points K, N to the
straight lines AB, AC, DE, DF, draw perpendiculars KB, KC, NJS;

NF ; and jom HB, BC, ME, EF : Because HK is perpendicular to
the plane BAC, the plane HBK which passes through HK is at right
angles (18. IL) to the plane BAC : and AB is drawn in the plane
BAC at right angles to the common section BK of the two planes]
therefore AB is perpendicular (4. def IL) to the plane HBK, and
makes right angles (3. def. IL) with every straight Ime meeting it
in that plane. But BH meets it in that plane ; therefore ABH is a
right angle. For the same reason, DEM is a right angle, and is
therefore equal to the angle ABH : and th6 angle HAB is equal to
the angle MDE. Therefore in the two triangles HAB, MDE there
are two angles in one equal to two angles in the other, each to each,
and one side equal to one side, opposite to one of the equal angles
in each, viz. HA equal to DM ; therefore the remaining sides are equal
(26. 1.) each to each : wherefore AB is equal to DE. In the same
manner, if HC and MF be joined, it may be demonstrated that AC

is equal to DF ; therefore, since AB is equal to DE, BA and AC are
equal to ED and DF ; and the angle BAC is equal to the angle EDF;
wherefore the base BC is equal (4. L) to the base EF, and the re-
maining angles to the remaining angles : the angle ABC is therefore

194 THE BLEMENTS OV EUCLID. . BOOK TLU

equal to the angle DBF: and the right angle ABK is equal to the
right angle DEN, whence the remaining angle CBK is equal to the
remaining angle FEN : for the same reason, the angle BCK is equal
to the angle EFN : therefore in the two triangles BCK, EFN there
are two angles in one equal to two angles in the other, each to each,
and one side equal to one side adjacent to the equal angles in each,
viz. BC equal to EF ; the other sides, therefore, are equal to the other
sides ; BK then is equal to EN ; and AB is equal to DE ; wherefore
AB, BK are equal to DE, EN ; and they contain right angles : where-
fore the base AK is equal to the base DN : and smce AH is equal to
DM, the square of AH is equal to the square of DM : but the squares
ci AK, KH are equal to the tquare (47. 1.) of AH, because AKH is
a right angle : and the squares of DN, NM are equal to the square
of DM, for DNM is a right angle : wherefore the squares of AK, KH
are equal to the squares of DN, NM ; and of those the square of AK
is equal to the square of DN ; therefore the remaining square of KH
IS equal to the remaining square of NM ; and the straight line KH to
the straight line NM ; and because HA, AK are equal to MD, DN,
each to each, and the base HK to the base MN, as has been proved;
therefore the angle HAK is equal (8. 1.) to the angle MDN. Q. E. D.
Cor. From this it is manifest, that if, from the vertices of two
equal plane angles, there be elevated two equal straight lines con-
taining equal angles with the sides of the angles^ each to each : the
perpendiculars drawn from the extremities of the equal straight lines
to the planes of the first angles are equal to one another.

Another Demonstration of the Corollary.

Let the plane angles BAC, EDF be equal to one another, and let
AH, DM, be two equal straight lines above the planes of the angles,
containing equal angles with BA, AC ; ED, DF, each to each, viz.
the angle HAB, equal to MDE, and HAC equal to the angle MDF ;
and from H, M let HK, MN be perpendiculars to the planes BAC,
EDF ; HK is equal to MN.

Because the solid angle at A is contained by the three plane an-
gles BAC, BAH, HAC, which are, each to each, equal to the three
plane angles EDF, EDM, MDF containing the solid angle at D ; the
solid angles at A and D are equal, and therefore coincide with one
another ; to wit, if the plane angle BAC be applied to the plane an-
gle EDF, the straight line AH coincides with DM, as was shown in
Prop. R of this Book : and because AH is equal to DM, the point H
coincides with the point M ; wherefore HK, which is perpendicular
to the plane BAC, coincides with MN (13. 1 1.)> which is perpendicu-
lar to the plane EDF, because these planes coincide with one another.
Therefore HK is equal to MN. d. E. D.

PROP. XXXVI. THEOR.

If three straight lines be proportionals, the solid paralleloprped
described from all three as its sides, is equal to the equilateral
parallelopiped described from the mean proportional, one of the

BOOK XI.

TUB BLfiMSNTH OF BIK3L1D.

solid angles of which is contained by three plane angles equal,
each to each, to the three plane angles containing one of the
solid angles of the other figure.*

Let A B, C, be three proportionals, viz. A to B, as B to C. The
solid described from A, B, C is equal to the equilateral solid de-
scribed from B, equiangular to the other. ,

Take a solid angle D contained by three plane angles EDF, PDG,
GDE ; and make each of the straight lines ED, DF, DG equal to B,
and complete the solid parallelepiped DH. Make LK equal to A,
and at the point K in the straight Kne LK make (26. 11.) a solid
angle contained by the three plane anf les LKM, MKN, NKL, equal
to the angles EDF, FDG, GDE, each to each ; and make KN equal
to B, and KM equal to C; and complete the solid parallelepiped

EO, and because, as A is to B, so is B to C, and that A is equal to
LK, and B to each of the straight lines DE, DF, and C to KM : there-
fore LK is to ED, as DF to KM ; that is, the sides about the equal
angles are reciprocally proportional ; therefore the parallelogram LM
is equal (14. 6.) to EF: and because EDF, LKM are two equal plane
angles, and the two equal straight lines DG, KN are drawn from
their vertices above their planes, and contain equal angles with their
sides ; therefore the perpendiculars from the points G, N, to the planes
EDF, LKM are equal (Cor. 35. 11.) to one another: therefore the
solids KO, DH are of the same altitude ; and they are upon equal
bases LM, EF, and therefore they are equal (31. 11.) to one another:
but the solid KO is described from the three straight lines A, B, C,
and the solid DH from the straight line B. If therefore three straight
lines, &c. Q. £L 0.

PROP. XXXVn. THEOR.

If four straight lines be proportionak, the similar solid paral-
lelepipeds similarly described from them ^all also be propor-
tionals. And if the similar paraUelopipeds similarly described
from four straight lines be proportionals, the straight lines shall
be proportionals.*

« See Note.

THB BLUIBNTS QT TOOLID,

BOOK XU

Let the four straight lines AB» CD, EF» OH be proportionals, viz.
as AB to CD, so £F to OH ; and let the similar paralldopipeds AK»
CL, EM, GN be similarly described from them* AK is to CL, as
EM to GN.

Make (11. 0.) AB, CD, O, P continual proportionals, as also EF,
GH, Q, R; and because as AB is to CD, so EF to GH; and that

JkC

CD is (11. 5.) to O, as GH to d, and O to P, as Q. to R; therefore,
ex seqiudi (22. 5.), AB is to P, as EF to R : but as AB to P, so (Cor.
33. 11.) is the solid AE to the solid CL; and as EF to R, so (Cor.
33. 11.) is the solid EM to the solid GN; therefore (11. 5.) as the
soh'd AK to the solid CL, so is the solid EM to the solid GN.

But let the solid AK be to the solid CL, as the solid EM to the
solid GN : the straight line AB is to ED, as EF to GH.

Take AB to CD, as EF to ST, and from ST describe (27. 11.)
a solid paralldopiped SV similar and similarly situated to either
of the solids EM, GN : and because AB is to CD, as EF to

P *

ST, and that from AB, CD the solid parallelopipeds AK, CL are
similarly described, and in like manner the solids EM, SV from the
straight lines EF, ST ; therefore AK is to CL, as EM to SV : but.

BOOK XI, TRB BLIMBNT9 OP mrCLID. 197

by the hypothesis, AK is to CL, as EM to GN ; therefore GN is equal
(9. 6.) to SV : but it is likewise similar and similarly situated to SV ;
therefore the planes which contain the solids GN, SV are similar and
equal, and their homologous sides GH, ST equal to one another :
and because as AB to CD so EF to ST, and that ST is equal to GH,
AB is to CD, a£i EF to GH. Therefore, if four straight lines, ^.
Q. R D.

PROP. XXXVffl. THEOR.

•

" If a plane be perpendicular to another plane, and a straight
line be drawn from a point in one of the planes perpendicular to
the other plane, this straight line shall fall on the common section
of the planes.''*

** Let the plane CD be perpendicular to the plane AB, and let AD
be their common section ; if any point E be taken in the plane CD,
the perpendicular drawn from B to the plane AB shall fall on AD.

" For, if it does not, let it, if possible, M elsewhere, as EF; and
let it meet the plane AB in the point F; and from F draw (12. 1.) in
the plane AB a perpendicular FG to DA, which is also perpendicular
(4. de£ 11.) to the plane CD; and jom EG: then because FG is per-
poidicular to the plane CD, and the C
straight line EG, which is in that plane,
meets it ; therefore FGE is a right an-
gle (a def 11.): but EF is also at right
angles to the plane AB ; and therefore .
EFG is a right angle : wherefore two
of the angles of the triangle EFG are
equal together to two right angles;
Which is absurd: therefore the perpen-
dicular from the point E to the plane AB, does not fall elsewhere than
upon the straight line AD ; it therefore fells upon it. If therefore a
plane," &c. Q, E. D.

PROP. XXXIX. THEOR.

Ik a solid parallelopiped, if the sides of two of the opposite
planes be divided each into two equal parts, the common section
of the planes passing through the points of division, and the di-
ameter of the solid parallelopiped cut each other into two equal
parts.*

* See Note.

198

THE KLiaMENTS OF BaCLID.

BOOK XI.

Let the sides of the D K F

opposite planes CF,
AH of the solid paral-
lelopiped AF, be di-
vided each into two
equal parts in the
pointsK, L, M, N;X,
O, P, R ; and join KL,
MN, XO, PR: and
because DK, CL are
equal and parallel, KL
is parallel (33. L) to
DC : for the same rea- B
son, MN is parallel to
BA : and BA is paral-
lel to DC; therefore
because KL, BA, are
each of them parallel to DC, and not in the same plane with it, KL
is parallel (9. H.) to BA: and because KL, MN are each of them pa-
rallel to BA, and not in the same plane with it, KL is parallel (9. IL)
to MN ; wherefore KL, MN are in one plane. In like manner, it may
be proved, that XO, PR are in one plane. Let YS be the common
section of the planes KN, XR ; and DG the diameter of the solid
parallelopiped AF : YS and DG do meet, and cut one another into two

equal parts.

Join DY, YE, BS, SG. Because DX is parallel to OE, the al-
ternate angles DXY, YOE are equal (29. L) to one another: and

because DX is equal
to OE, and XY to YO,
and contain equal an-
gles, the base DY is
equal (4. 1.) to the
base YB, and the
other angles are equal ;
therefore the angle
XYD is equal to the
angle OYE, and DYE
is a straight (14. L)
Mne : for the same rea-
son fiSG is a straight
Hne, and BS equal to
SG : and because CA
is equal and parallel
to DB, and also equal
and parallel to EG,

therefore DB is equal and parallel (9. 11.) to EG: and DE, EG join
their extremities ; therefore DE is equal and parallel (33. 1.) to BG:
and DG, YS are drawn from points in the one, to points fai the other:
and are therefore in one plane : whence it is manifest, that DG, YS
must meet one another ; let them meet in T : and because DE is pa-
rallel to BG, the alternate angles EDT, BGT are equal (29. 1.);
and the angle DTY is equal (16. 1.) to the angle GTS : therefore in

BOOK XI.

THJ3 ELfiMWTS OF BUGLID.

199

the triangles DTY, GTS there are two angles in the one equal to
two angles in the other, and one side equal to one side, opposite to
two of the equal angles, viz. DY to.GS ; for they are the halves of
DK, 6G: therefore the remaining sides are equal (26. 1.) each to
each. Wherefore, DT is equal to TG, and YT equal to TS. Where-
fore, if in a solid, &c. Q,. E. IX

PROP. XL. THEOR.

If there be two triangular prisms of the same altitude, the base
of one of which is a parallelogram, and the base of the other a
triangle; if the parallellogram be Vouble of the triangle, the
prisms shall be equal to one another.

Let the prisms ABCDEF, GHELMN be of the same altitude, the
first whereof is contained by the two triangles ABE, CDF, and the
three parallelograms AE, DE, EC ; and the other by the two trian-
gles GHK, LMN, and the three parallelograms LH, HN^ NG ; and
let one of them have a parallelogram AF, and the other a triangle
GHK for its base ; if the paralldogram AF, be double of the triangle
GHK, the prism ABCDEF is equal to the prism GHKLMN.

Complete the solids AX, GO; and because the parallelogram AF
is double of the triangle GHK ; and the paraUelogram HK double

A\Hr-

Jc\

(34. I.) of the same triangle ; therefore the parallelogram AF is equal
to HK. But solid parallelepipeds upon equal bases, and of the
same altitude, are equal (31. IL) to one another. Therefore the
solid AX is equal to the solid GO ; and the prism ABCDEF is half
(28. 11.) of the solid AX ; apd the prism GHKLMN half (28. 11.) of
the solid GO. Therefore the prism ABCDEF is equal to the prism
GHKLMN. Wherefore, if there be two, &c. Q. E. D.

THE

ELEMENTS OF EUCLID.

BOOK XII.

LEMMA I.

Which is the first proposition of the tenth book, and is necessary to
some of the propositions of this book.

If from the greater of two unequal magnitudes, there be taken
more than its half, and from the remainder more than its half,
and so on : there shall at length remain a magnitude less than
the least of the proposed magnitudes.*

Let AB and C be two unequal magnitudes, of
which AB is the greater. If from AB there be taken
more than its half, and firom the remainder more D

than its half, and so on ; there shall at length re-
main a magnitude less than C.

For C may be multiplied, so at length to become
greater than AR Let it be so multiplied, and let DE
its multiple be greater thsui AB, and let DE be di- „ p- -

vided into DF, FG, GE, each equal to C From AB
take BH greater than its half, and from the remain-
der AH take HK greater than its half, and so on,
until there be as many divisions in AB as there are
in DE : and let the divisions in AB be AK, KH, HB;
and the divisions in ED be DF, FG, GE. And be-
cause DE is greater than AB, and that EG taken B C
from DE is not greater than its half, but BH taken from AB is great-
er than its half; therefore the remainder GD is greater than the re-
mainder HA. Again, because GD is greater than HA, and that GF
is not greater than the half of GD, but HK is greater than the half
of HA ; therefore the remainder FD is greater than the remainder

G—

* See Note.

BOOK xn.

THE BLfiMVirrS Of BVGIID.

AK And FD is equal to C, therefore C \m greater than AK ; that ia,
AK is less than C. Q. £. D.

And if only the halves be taken away, the same thing may in the
same way be demonstrated.

PROP. I. THBOR.

Similar polygons inscribed in circles are to one another as the
squares of their diameters.

Let ADCDE, FGHKL be two circles, and tn them the simOat poly-
gons ABODE, FGHKL ; and let BM, GN be the diameters of the
circles ; as the square of BM is to the square of GN, so is the poly-
gon ABCD£ to the polygon FGHKL.

Join BE, AM, GL, FN : and because the polygon ABODES is simi-
lar to the polygon FGHKL, and similar polygons are divided into
similar triangles : the triangles ABE^ FGL are similar and equiango-

A F

lar (6. 6.) ; and therefore the angle A£B is equal to the angle FLQ:
but AEB is equal (21. 3.) to AMB, because they stand upon the
same circumference ; and the angle FLG is, for the same reason, equal
^ to the angle FNG : therefore also the angle AMB is equal to FNG :
' and the right angle BAM is equal to the right (31. 3.) angle GFN;
wherefore the remaining angles in the triangle ABM, FGN are equal*
and they are equiangular to one another : therefore as BM to GN, so
(4. 6.) is BA to GF ; and therefore the duplicate ratio of BM to GN,

A F

is the same (10. def. 5. and t% 6.) with the duplicate ratio of BA to
GF : but the ratio of the square of BM to the square of GN is the
duplicate (20. 6.) ratio of that which BM has to GN ; and the ratio of
the polygon ABCD^ to the polygon FGHKL is the duplicate (30. 6.)
of that which BA has to GF: therefore, as the square of BM to the

THfi ELEMKNTB OF EUCLID. BOOK XII.

square of ON, so is the polygon ABODE, to the polygon FGHKL.
Wherefore, similar polygons, &c. Q. E. D.

PROP. n. THEOR.

Circles are to one another as the squares of their diameters.*

Let ABCD, EPdH be two circles, and BD, FH their diameters: as
the square of BD to the square of FH, so is the circle AQCD, to the
circle EFGH.

For, if it be not so, the square of BD shall be to the square of FH,
as the circle ABCD is to some space either less than the circle EFGH,
or greater than it.t First, let it be to a space S less than the circle
EFGH ; and in the circle EFGH describe the square EFGH : this
square is greater than half of the circle EFGH : because i( through
the points E, F, G, H, there be drawn tangents to the circle, the
square EFGH is half (41. 1.) of the square described about the circle;
and the circle is less than the square described about it ; therefore
the square EFGH is greater than half of the circle. Divide the cu:-
cumferences EF, FG, GH, HE, each into two equal parts in the points
K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE : therefore
each of the triangles EKF, FLG, GMH, HNE is greater than half of
the segment of the circle it stands in ; because, if straight lines touch-
ing the circle be drawn through the points K, L, M, N, and parallelo-
grams upon the straight lines EF, FG, GH, HE be completed ; each
of the triangles EKF, FLG, GMH, HNE shall be the half (41. 1.)
of the parallelogram in which it is : but every segment is less than
the parallelogram in which it is: wherefore each of the triangles
EKF, FLG, GMH, HNE is greater than half the segment of the cir-
cle which contains it : and if these circumferences before named be
divided each into two equal parts, and their extremities be joined

by straight lines, by continuing to do this, there wfll at length r^
main segments of the circle, which, together, shall be less than the

« SpeNote.

t For there is some square equal to the circle ABCD; let. F be the side of it;
and to three straight lines BD, FH, and P, there can be a fourth proportional ; let
this lie Q : therefore the squares of these four straight lines are proportionals, that
is, to the squares oT BD, FH, and the circle ABCD, it is possible there may be a
fourtii proportional. Let this be S» And in like mamier, are to be understood
■ome things in some of the following propositions.

BOOK XU,

TRB BLBMENT9 OP EOCLID.

AAA

excess of the circle EFGH above the space S : because, by the pre^
ceding lemma, if from the greater of two unequal magnitudes there
be taken more than its half, and from the remainder more than its
half, and so on, there shall at length remain a magnitude less than
the least of the proposed magnitudes. Let then the segments EK, KF,
FL, LG, GM, MH, HN, NE, be those that remain and are together
less than the excess of the circle EFGH above S : therefore the rest
of the circle,, viz. the polygon EKFLGMHN, is greater than the space
S. Describe likewise in the circle ABCD the polygon AXBOCPDR
similar to the polygon EKFLGMHN: as therefore, the square
of BD is to the square of FH, so (1. 12.) is the polygon AXBOCPDR
to the polygon EKFLGMHN : but the square of BD is also to the
square of FH, as the circle ABCD is to the space S : therefore as the
circle ABCD is to the space S, so is (IL 5.) the polygon AXBOCPDR
to the polygon EKFLGMHN : but the circle ABCD is greater than
the polygon contained in it : wherefore the space S is greater (14. 5.)

R Z^

than the polygon EKFLGMHN : but it is likewise less, as has been
demonstrated : which is impossible. Therefore the square of BD is
not the square of FH, as the circle ABCD is to any space less than
the circle EFGH. In the same manner it may be demonstrated,
that neither is the square of FH to the square of BD, as the circle
EFGH is to any space less than the circle ABCD. Nor is the square
of BD to the square of FH, as the circle ABCD is to any space
greater than the circle EFGH : for, if possible, let it be so to T, a
space greater than the circle EFGH : therefore, inversely, as the
square of FH to the square of BD, so is the space T to the circle
ABCD. But as the space T* is to the circle ABCD, so is the curcle

* For, as in the foregoing note, at t it was explained how it wad possible there
could be a fourth proportional to the squares of BD, FH, and the circle ABCD,
which was named S. So in like manner there can be a fourth proportional to this
other space named T, and the circles ABCD, EFGH. And the like is to be
understood in some of the following propositions.

THB BLnfBim OF BUCUD.

BOOK XII.

KFQH to some space, which must be less (14. 5.) th«n the circle
ABCD« because the space T is greater by hypothesis, than the circle
BFQH. Therefore as the square of FH is to the square of BD, so
is the circle EFGH to a space less than the circle ABOD, which has
been demonstrated to be impossible : therefore the square o( ED is
not to the square of FH, as the circle ABCD is to any space greater
than the cude EFGH : and it has been demonstrated, that neither is
the square of ED to the square of FH, as the circle ABCD is to any
space less than the circle EFGH : wherefore, as the square of BD to
the square of FH, so is the circle ABCD to the circle EFGH * Cir-
cles therefore are, &c. CI. E. D.

PROP. HI. THEOR.

Every pyramid having a triangular base, may be divided into
two equal and similar pyramids having triangular bases, and
which are similar to the whole pyramid ; and into two equal
prisms which together are greater than half of the whole pyra-
mid.t

Let there be a pyramid of which the base is the triangle ABC,
and its vertex the point D : the pyramid ABCD may be divided into
two equal and similar pyramids having triangu^ ^

lar bases, and similar to the whole; and into
two equal prisms which together are greater
than half of the whole pyramid.

Divide AB, BC, CA, AD, DB, DC, each into
two equed parts in the points E, F, G, H, E, L,
and join EH, BG, GH, HE, EL, LH, EE, EF,
FG. Because AE is equal to EB, and AH to
HD, HE is parallel (2. 6.) to DB ; for the same
reason, HE is parallel to AB : therefore HEBE
is a parallelogram, and HE equal (34. 1.) to ^B:
butEBis equal to AE; therefore also AE is
equal to HE : and AH is equal to HD ; where-
fore EA, AH are equal to EH, HD, each to
each; and the angle EAH is equal (29. 1.) to ^
the angle EHD ; therefore the base EH is equal
to the base ED, and the triangle AEH equal (4. L) and similar to
the triangle HEQ : for the same reason, the triangle AGH is equal
and similar to the triangle HLD : and because the two straight
lines EH, HG, which meet one another, are parallel to KD, DL,
that meet one another, and are not in the ssune plane with them,
they t contain ec^ual (10. U.) angles; therefore the angle EHG is
equal to the angle EDL. Again, because EH, HG, are equal to
ED, DL each to each, and the angle EHG equal to the angle EDL ;
therefore the base EG is equal to the base EL ; and the triangle

* Because as a fourth proportional to the squares of BD, FH, and the circle
ABCD is possible, and that it can neither be less nor £;reater than the circle EF^H,
it roust be equal to it

t See Note.

ncuD. W6

■EHa equal (4. I.) and aimllar to the triangks KDL ; for the Bame
reason the triangle AEQ is also equal and similar to the triangle
HKL. Therefore the pyramid of which the base is the triangle ABG,
and of which the vertex is the point H, is equal (C. 11.) and similar
to the pyramid the base of which is the triangle KHU and vertex
the point D : and ttecause HK is parallel to AB a D

side of the trian^e ADB, the triangle ADB, is
eqidangular to the triai^le HDB^ and their sides
are prc^rtionala (4. S.): therefore the triangle
ADB is similar to the triangle HDK : and for the
same reason, the triangle DEC is similar to the
triangle DKL; and the triangle ADC to the tri-
angle HDL ; and also the triangle ABC to the
triangle AEO : but the triangle AEG is similar to
the triangle HKL, as before was proved ; there-
fore the triangle ABC Is similar (21. 6.) to the
triangle HKL. And the pyramid of which the
base is the triangle ABC, and vertex the point D,
is therefore similar (B Il.andll.def. 11.) to the i
pyramid of which the base is the triangle HKL, t
and vertex the same point D; but the pyramid B F C

of which the base is the triangle HKL, and vertex the point D, is
similar, as has been proved, to the pyramid the base of which is the
trian^ ABG, and vertex the point H : wherefore the pyramid, the
base of which is the triangle ABC, and vertex the point D, is similar
to the pyramid of which the base is the triangle AEG and vertex H:
therefore each of the pyramids AEGH, HKLD is similar to the whole
pyramid ABCD: and because BF is equal to FC, the parallelogram
EBFG ia double (41. 1.) of the triangle GFC : but when there are
two prisms of the same altitude, of which one has a paralleI(^T&n]
for its bass, snd the other a triangle that is half of the parallelogram,
these i»isms are equal (40. 1 1.) to one another ; therefore the prism
having the parallelogram EBFG for its base, and the strai^t line
KH opposite to it, is equal to the prfsm having the triangle GFC for
its base, and the triangle HKL opposite to it ; for they are of the
same altitude, because they are between the parallel (IS. 11.) planes
ABC, HKL : and it is manifest that each of these prisms is greater
than either of the pyramids of which the triangles AEG, HKL are
the liases, and the vertices the points H, D ; because if E, F be joinr-
ed, ttie prism having the parallelogram EBFG for its base, and KH
the straight line opposite to it, is greater than the pyramid of which
the base ia the triangle EBF, and vertex the point K ; but this pyra-
mid ia equal (C. 1 1 .) to the pyramid the base of which is the triangle
AEG, and vertex the point H ; because they are contained by equal
and similar planes: wherefore the prism having the parall^ogram
EBFG for its base, and opposite side KH, is greater than the pyramid
of which the base la the triangle AEG, and vertex the point H : and
the prism of which the base is the parallelogram EBFG, and oppo-
site side EH, fs equal to the prism having the triangle GFC for its

THB ELEMENTS OF EUCLID. BOOK XII.

base, and HKL the triangle opposite to it ; and the pyramid of which*
the base is the triangle AEG, and vertex H, is equal to the pyramid
of which the base is the triangle HKL, and vertex D : therefore the
two prisms before mentioned are greater than the two pyramids of
which the bases are the triangles AEG, HKL, and vertices the points
H, D. Therefore the whole pyramid of which the base is the trian-
gle ABC, and vertex the point D, is divided into two equal p3nramids
similar to one another, and to the whole pyramid; and into two
equal prisms ; and the two prisms are together greater than half of
the whole pyraniid. Q. E. D.

PROP. IV. THEOR.

If there be two pyramids of the same altitude, upon triaogular
bases, and each of them be divided into tvtro equal pyramids
similar to the whole pyramid, and also into two equal prisms ;
and if each of these pyramids be divided in the same manner as
the first two, and so on : as the base of one of the first two
pyramids is to the base of the other, so shall all the prisms in one
of them be to all the prisms in the other, that are produced by
the same number of divisions.*

Let there be two pyramids of the same altitude upon the trian-
gular bases ABC, DEF, and having tiieir vertices in the points G,
H ; and let each of them be divided into two equal pyramids similar
to the whole, and into two equal prisms ; and let each of the pyra^
mids^thus made be conceived to be divided in the like manner, and
80 on : as the base ABC is to the base DEF, so are all the prisms in
the pyramid ABCG, to all the prisms in the pyramid DEFH made by
the same number of divisions.

Make the same construction as in the foregoing proposition : and
because BX is equal to XC, and AL to LC ; therefore XL is paral-
lel (2. 6.) to AB, and the triangle ABC similar to the triangle LXC :
for the same reason, the triangle DEF is similar to RVF : and be-
cause BC is double of CX, and EF double of FV, therefore BC is to
CX, as EIF to FV : and upon BC, CX are described the similar and
similarly situated rectUineal figures ABC, LiXC ; and upon EF, FV,
in like manner, are described the similar figures DEF, RVF: there-
fore, as the triangle ABC is to the triangle LXC, so (22. 6.) is the
triangle DEF to the triangle RVF, and, by permutation, as the trian-
gle ABC to the triangle DEF, so is the triangle LXC to the triangle
RVF : and because the planes ABC, OMN, as also the planes DEF,
STY are parallel (5. 11.), the perpendiculars drawn from the points G,
H to the bases ABC, DEF, which, by the hypothesis, are equal to one
another, shall be cut each into two equal (17. 11.) parts by the planes
OMN, STY, because the straight lines GO, HFare cut into two equal

« See Note.

BOOK XIL

TBI BLEHENTS Of EUCLID.

207

parts in the points N, Y by the same planes: therefore the prisms
LXCOMN, RVFSTY are of the same altitude; and therefore as the
base LXC to the base RVF ; that is, as the triangle ABC to the tri-
angle DEF, so {Cor. 32. 11.) is the prism having the triangle LXC
for its base, and OMN the triangle opposite to it, to the prism of which
the base is the triangle RVF, and the opposite triangle STY : and b&.
cause the two prisms in the pyramid ABCG are equal to one another,
and also the two prisms in the pyramid DEPH equal to one another,
as the prism of which the base is the parallelogram KBXL and oppo-
site side MO, to the prism having the triangle LXC for its base, and
OMN the triangle opposite to it, so is the prism of which the base
(7. 5.) is the parallelogram PEVR, and opposite side TS. to the prism
of which the base is the triangle UVP, and opposite triangle STY.
Therefore componendo, as the prisms KBXLMO, LXCOMN together
are unto the prism LXOMN, so are the prisms PEVRTS, RVFSTY,
G H

E V F

to the prism RVFSTY; and permutando, as the prisms KBXLMO,
LXCOMN are to the prisms PEVRTS, RVFSTY, so is the prism
LXCOMN to the prism RVFSTY: but as the prism LXCOMN to
the prism RVFSTY, so is, as has been proved, the base ABC to the
base DEF : therefore, as the base ABC to the base DEF, so are the
two prisms in the pyramid ABCG to the two prisms in the pyramid
DEFH r and likewise if the pyramids now made, for example, the two
OMNG, STYH, be divided in the same manner; as the base OMN
b to the base STY, so shall the two prisms in the pyramid OMNG
be to tlie two prisms in the pyramid STYH : but the base OMN is
to the base STY, as the base ABC to the base DEF ; therefore, as
the base ABC to the base DEF, so are the two prisms in the pyra-
mid ABCG to the two prisms in the pyramid DEFH; and so are the
two prisms in the pyramid OMNG to the two prisms in the pyramid
STYH ; and so are all four to ali four : and the same thing may be
showa of the prisms made by dividing the pyramids AKLO and
DPRS, and of all made by the same number of divisions. ^ E, D.

B MiiKMMtfn or rucuD.

PROP. V. THEOR.

PnuMiDs of the same altitude, which have triangular bases,
are to one another as their bases.*

Let the pyramida of which the triangles ABC, DEF are the bases,
and of which the vertices are the points G,H, be of the same altitude;
as the base ABC, to the base DEF, so fs the pyramid ABCG to the
pyramid DEFH.

For, if it be not so, the base ABC must be to the base DEF, as the
pyramid ABCG to a solid either less than the pyramid DEFH, or
greater than itt First, let it be to a solid less than it, viz. to the
solid Q,: and divide the pyramid DEFH into two equal pyramids,
similar to the whole, and into two equal prisms : therefore these two
prisms are greater (3. 12.) than the half of the whole pyramid. And
again, let the pyramids made by this division be in like manner di-
vided, and so on, until the pyramids which remain undivided in tbe
pyramid DEFH be. all of them together, less than the excess of the
pyramid DEFH above the solid Q, : let these, for example, be the
pyramids DPBS, ST YH : therefore the prisms, which make the rest
of tbe pyramid DEFH, are greater than the solid Q; divide likewise
the pyramid ABCG in the same manner, and into as many parts, as
the pyramid DEFH : therefore, as the base ABC to the base DEF, so
(4. 12.) are tbe prisms in the pyramid ABCG to the prisma in the
pyramid DEFH: but as the base ABC to the base DEF, so, by hy-
pothesis, is the pyramid ABCG to the solid Q, ; and therdbre, as the
pyramid ABCG to the solid Q, so are the prisms in the pjrramid
ABCG, to the prisms in the pyramid DEFH ; but the pyramid ABCG
is greater than the prisnts contained in it ; wherefore (14. 5.) also

G H

it the Dute i ia propoiitioD 9 in tbe

BOOK XII. THE ELBMBHTS OF EGCUD, W^

the solid Q, is greater than the prisms in the pyramid DEFH. But
it is also less, which is impossible. Therefore the base ABC is not
to the base DEF, as the pyramid ABCG to any solid which is less
than the pyramid DEFH. In the same manner it may be demon-
strated, that the base DEF is not to the base ABC, as the pyramid
DEFH to any solid which is less than the pyramid ABCG. Nor can
the base ABC be to the base DEF, as the pyramid ABCG to any
solid which is greater than the pyramid DEFH. For, if it be possi-
ble, let it be so to a greater, viz. the solid Z. And because the base
ABC is to the base DEF, as the pyramid ABCG to the solid Z ; by
inversion, as the base DEF to the base ABC, so is the solid Z to the
pyramid ABCG. But as the solid Z is to the pyramid ABCG, so is
the pyramid DEFH to some solid,* which must be less (14. 5.) than
the pyramid ABCG, because the solid Z is greater than the pyramid
DEFH. And therefore, as the base DEF to the base ABC, so is the
P3nramid DEFH to a solid less than the pyramid ABCG; the contrary
to which has been proved. Therefore the base ABC is not to the
base DEF, as the p3rramid ABCG to any solid which is greater than
the pjrramid DEFH. And it has been proved, that neither is the
base ABC to the base DEF, as the P3rramid ABCG to any solid
which is less than the pyramid DEFH. Therefore, as the base ABC
is to the base DEF, so is the pyramid ABCG to the pyramid DEFH.
Wherefore pyramids, &c. €t. E. D.

PROP. VI. THEOR.

PYRAiaDs of the same altitude, which have polygons for their
bases, are to one another as their bases.t

Let the pyramids which have the polygons ABCDE, FGHKL for
their bases, and their vertices in the points M, N, be of the same al-
titude : as the base ABCDE to the base FGHKL, so is the pyramid
ABCDEM to the pyramid FGHKLN.

Divide the base ABCDE into the triangles ABC, ACD, ADE ; and
the base FGHKL into the triangles FGH, FHK, FKL : and upon the
bases ABC, ACD, ADE let there be as many pyramids of which the
common vertex is the point M, and upon the remaming bases a»
many pyramids having their common vertex in the point N : th^e^

M N

B C G H

» This may be explained the same way as the Kke at the mark tin piop;^

t See Note.

91-0 THB ELEMBNTS OF EUCLID. BOOK Xll

fore, since the triangle ABC is to the triangle FGH, as (5. 12.) the
pyramid ABCM to the pyramid FGHN ; and the triangle ACD to the
triangle FGH, as the pyramid ACDM to the pyramid FGHN ; and
also the triangle ADE to the triangle FGH, as the pyramid ADEM
to the pyramid FGHN ; as all the first antecedents to their commcm
consequent, so (2. Cor. 24. 5.) are all the other antecedents to their
common consequent : that is as the base ABCDE to the base FGH,
' go is the pyramid ABCDEM to the pjrramid FGHN : and, for the
same reason, as the base FGHEL to the base FGH, so is the pyra-
mid FGHKLN to the pjrramid FGHN : and, by inversion, as the
base FGH to the base FGHKL ; so is the pyramid FGHN to the
pyramid FGHKLN : then, because as the base ABCDE to the base
FGH, so is the pyramid ABCDEM to the pyramid FGHN ; and as
the base FGH to the base FGHKL; so is the pyramid FGHN to
the pyramid FGHKLN ; therefore, ex sequali, (22. 5.) as the base
ABCDE to the base FGHKL, so the pyramid ABCDEM to the pyra-
mid FGHKLN. Therefore, pjrramids, &c. Q. E. D.

PROP. Vn. THEOR.

Every prism having a triangular base, noay be divided icibo
three pyramids that have triangular bases, and are equal to one
another.

Let there be a prism of which the base is the triangle ABC» and
let DEF be the triangle opposite to it : the prism ABCDEF may be
divided into three equal pyramids having triangular bases.

Join BD, EC, CD ; and because ABED is a parallelogram of which
BD is the diameter, the triangle ABD is equal (34. L) to the triangle
EBD ; therefore the pyramid of which the base is the triangle ABD,
and vertex the point C, is equal (5. 12.) to the pyramid of which the
base is the triangle EBD, and vertex the point C ; but this pyramid
is the same with the pyramid the base of which is the triangle EBC,
and vertex the point D ; for they are contained by the same planes :
therefore the pyramid of which the base is the triangle ABD, and
vertex the point C, is equal to the P3rramid, the base of which is the
triangle EBC, and vertex the point D : again, because FCBE, is a
parallelogram of which the diameter is CE, the F

triangle ECF is equal (34. 1.) to the triangle
ECB ; therefore the pyramid of which the base
is the triangle ECB, and vertex the point D, is D
equal to the pyramid, the base of which is the
triangle BCP, and vertex the point D : but the
pyramid of which the base is the triangle ECB,
and vertex the point D, has been proved equal
to the pyramid of which the base is the trian-
gle ABD, and vertex the point C. Therefore A B
the prism ABCDEP is divided into three equal pyramids having
triangular bases, viz. into the pyramids ABDC, EBDC, ECFD : and

BOOK XII.

THE BUHCENTS OF EVCLIO.

811

because the pyramid of wt^ch the base is the tiiaogle ABD« and ver*
tex the point C, is the same with the pyramid of which the base is
the triangle ABC, and vertex the point D, for they are, contained
by ti)e same planes; and that the pyramid of which the base is the
triangle ABD, and vertex the point C, has been demonstrated to bt
a third part of the prism, the base of which is the triangle ABC, and
to which DEF is the opposite triangle; therefore the pyramid of which
the base is the triangle ABC, and vertex the point D, is the third part
of the prism which has the same base, viz. the triangle ABC, and
DEF is the opposite triangle, d. E. D.

CoR. 1. From this it is manifest, that every pyramid is the third
part of a prism which has the same base, and is of an equal altitud?
with it ; for if the base of the prism be any other figure than a trianr
gle, it may be divided into prisms having triangular bases.

CoR. 2. Prisms of equal altitudes ^e to one another as their bases;
because the pyramids upon the same bases, and of the same altitude,
are (6. 12.) to one another as their bases.

PROP. Vm. THEOR.

SiKii^AR pyramids having triangular bases are one to another
in ihe triplicate ratio of that of their homologous sides.

Let the pyramids having the triangles ABC, DEF for their bases,
and the points G, H for their vertices, be similar, and similarly situ-
ated ; the pyramid ABCG has to the pyramid DEFH, the triplicate
ratio of that which, the side BC has to the homologous side EF.

Complete the parallelograms ABCM, GBCN, ABGK, and the solid
par^llelopiped BGML contained by these planes and those opposite to

them : and, in like manner, complete the solid paralleloplped EHPO con-
tained by the three parallelograms DEFP, HEFR, DEHX, and those
opposite to theto : and, because the pyramid ABCG is similar to the py-
ramid DEFH, the angle ABC is equal (11. def. 11.) to the angle DEF
and the angle GBC to the angle HEF, and ABG to DEH : and AB is

tl2 THB SLGMENTfi OP IVOLID. BOOK XII.

(1. def. 6.) to BO as DE to EF; that is, the sides about the equal angles
are proportionals; wherefore the parallelogram BM is similar to EP :
for the same reason, the parallelogram BN is similar to ER, and BK
to EX; therefore the three parallelograms BM, BN, BK are similar to
the three EP, ER, EX ; but the three BM, BN, BE, are equal and si-
milar (24.- 1 1.) to the three which are opposite to them, and the three
£P, ER, EX, equal and simUar to the three opposite to them : where-
fore the solids BGML, EHPO are contained by.the same number of
similar planes ; and their solid angles are equal (B. 1 1.) ; and therefore
the solid BGML, is similar (11. def. IL) to the solid EHPO : but simi-
lar solid parallelopipeds have the triplicate (33. 11.) ratio of that which
their homologous sides have : therefore the solid BGML has to the
solid EHPO the triplicate ratio of that which the side BC has to the
homologous side EF : but as the solid BGML is to the solid EHPO,
so is (15. 5.) the pyramid ABCG to the pyramid DEFH ; because the
P3rramtds are the sixth part of the solids ; since the prism, which is the
half (28l 11.) of the solid parallelopiped is triple (7. 12.) of the pyra-
mid. Wherefore likewise the pyramid ABCG has to the pyramid
DEFH the triplicate ratio of that which BC has to the homologous
side EF. a. E. D. .

CpR. From this it is evident, that similar pyramids which haye
multangular bases, are likewise to one another in the triplicate ratio
of their homologous sides: for they may be divided into similar pjrra-
mids having triangular bases, because the similar polygons, whicb
are their bases, may be divided into the same number of similar tri-
angles homologous to the polygons ; therefore as one of the triangu-
lar pyramid in the first multangular pyramids is to one of the trian-
gular pyramids in the other, so are all the triangular pyramids in the
first to all the triangular pyramids in the other ; that is, so is the first
multangular pyramid to the other : bqt one triangular pyramid is to
its similar triangular pyramid, in the triplicate ratio of their homolo-
gous sides ; and therefore the first multangular pyramid has to the
other, the triplicate ratio of that which one of the sides of the first has
to the homologous, side of the other.

PROP. IX. THEOR.

The bases aqd altitudes of equal pyramids having triangular
bases are reciprocally proportional : and triangular pyramids of
which the bases and altitudes are reciprocally proportional, are
equal to one another.

Let the pyramids of which the triangles ABC, DEF are the bases,
and which have their vertices in the points G, H, be equal to one
another: the bases. and altitudes of the pyramids ABCG, DEFH are
reciprocally proportional, viz. the base ABC is to the base DEF, as
the altitude of the pyramid DEFH to the altitude of the pyramid
ABCG.

Complete the parallelograms AC, AG, GC, DF, DH, HP, and

BOOK ZII.

THE BLBMBNTS OF BITCLtD.

21S

the solid fparalleloplpeds B6ML, EHPO contained by these planes
and those opposite to them : and because the pyramid ABCG is equal
to the pyramid DEFH, and that the solid BGML is sectuple of the
pyramid ABCG, and the solid EHPO sectuple of the pyramid DEFH ;
therefore the solid BGML is equal (1. Ax. 5.) to the solid EHPO :
but the bases and altitudes of equal solid parallelopipeds are recipro-
cally proportional (34. 11.); therefore as the base BM to the base
EP, so is the altitude of the solid EHPO to the altitude of the solid

BGML: but as the base BM to the base EP, so is (15. 5.) the trian-
gle ABC to the triangle DEF ; therefore as the triangle AJBC to the
triangle DEF, so is the altitude of the solid EHPO to the altitude of
the solid BGML : but the altitude of the solid EHPO is the same with
the altitude of the pyramid DEFH ; and the altitude of the solid
BGML is the same with the altitude of the pyramid ABCG : there-
fore, as the base ABC to the base DEF, so is the altitude of the
pyramid DEFH to the altitude of the P3nramid ABCG : wherefore the
bases and altitudes of the pyramids ABCG, DEFH are reciprocally
proportional.

Again, let the bases and altitudes of the P3a'amids ABCG, DEFH
be reciprocally proportional, viz. the base ABC to the base DEF, as
the altitude of the pyramid DEFH to the altitude of the pyramid
ABCG : the pyramid ABCG is equal to the pyramid DEFH.

The same construction being made, because as the base ABC to
the base DEF, so is the altitude of the pyramid DEFH to the altitude
of the pyramid ABCG : and as the base ABC to the base DEF, so is
the parallelogram BM to the parallelogram EP : therefore the paral-
lelogram BM is to EP, as the altitude of the pyramid DEFH to the
altitude of the pyramid ABCG : but the altitude of the pyramid DEFH
is the same with the altitude of the solid parallelopiped EHPO : and
the altitude of the pyramid ABCG is the same with the altitude of
of the sotid paraUelopiped BGML : as, therefore, the base BM to the
base EP, so is the altitude of the solid parallelopiped EHPO to the
altitude of the solid parallelopiped BGML. But solid parallelopipeds
having their bases and altitudes reciprocally proportional, are equal
(34. 11.) to one another. Therefore the solid parallelopiped BGML
Is equal to the solid parallelopiped EHPO. And the pyramid ABCG

214 THK KLBMBirra OF BUCLID. BOOK ZU.

is the sixth part of the sdid B6ML, and the pjrramid DBF^ is tlie
sixth part of the solid EPHO. Therefore the pyramid ABCD is equal
to the pyramid DEFH. Therefore the bases, d>c. Q. E. D.

PROP. X. THEOR,

EvEAY cone is a third part of a cylinder which has the same
base, and is of an equal altitude with it

Let a cone have the same base with a cylinder, viz. the circle
ABCD, and the same altitude. The cone is the third part of the
cylinder ; that is, the cylinder is triple of the cone.

If the cylinder be not triple of the cone, it must either be greater
than the triple, or less* than it. First, let it be greater than the triple :
and describe the square ABCD in the circle ; this square is greater
than the half of the circle ABCD :* Upon the square ABCD erect a
prism of the same altitude with the cylinder ; this prism is greater
than half of the cylinder : because if a square be described about the
circle, and a prism erected upon the square, of the same altitude
with the cylinder, the inscribed square is half of that circumscribed ;
and upon these square bases are erected solid parallelepipeds, viz.
the prisms of the same altitude ; therefore the prism upon the square
ABCD is half of the prism upon the square described about the cir-
cle: because they are to one another as their beises (32. 11.); and
the cylinder is less than the prism upon the square described about
the circle ABCD : therefore the prism upon the square ABCD of the
same altitude with the cylinder, is greater than half of the cylinder.^
Bisect the circumferences AB, BC, CD, DA in the points E, F, G, H
and join AE, EB, BF, FC, CG, GD, DH, HA : then each of the trian.
gles AEB, BFC, CGD, DHA is greater than half of the segment of the
circle in which it stands, as was sht»wn in prop. 2. of this book.
Erect prisms upon each of these triangles, of the same altitude
with the cylinder ; each of these prisms A

is greater than half of the segment of
the cylinder in which it is ; because if
through the points E, 1^, G, H, parallels
be drawn to AB, BC, CD, DA, and pa-
ralldograms be completed upon the
same AB, BC, CD, DA, and solid pa-
rallelopipeds be erected upon the pa-
rallelograms ; the prisms upon the tri-
angles AEB, BFC, CGD, DHA are the
halves of the solid paraMopipeds (2. C

Cor. 7. 12.). And the segments of the cylinder which are upon
the segments of the circle cut off by AB, BC, CD, DA, are less
than the solid parallelepipeds which contain them. Therefore the
prisms upon the triangles AEB, BFC, CGD, DHA, are greater
than half of the segments of the cylinder in which they are ; there-

* As was shown in prop. S. of this book.

BOOK Xll»

TBS BLBIUSKT8 OF fiOGLlD.

Sift

fore if each of the drcumfereoces be divided into two equal parts,
and straight lines be drawn from the points of division to the ex-
tremities of the chrenmferenoes, and upon the triangles thus made,
prisms be erected of the same altitude with the cylinder, and so on,
there must at length remain some segments of the cylinder which
together are less (Lem.) than the excess of the cylind^ above
the triple of the cone. , Let them be those upon the s^ments of the
circle AE, EB, BF, FC, CG, GD, DH, HA. Therefore the rest of the
cylinder, that is, the prism of which the base is the polygon
ABBFCGDH, and of which the altitude is the same with that of the
cylinder, is greater than the triple of the cone: but this prism is triple
(1. Cor. 7. 12.) of the pyramid upon the same base, of which the ver-
tex is the same with the vertex of the cone ; therefore the pyramid
upon the base AEBFCGDH, having the same vertex with the eone,
» greater than the cone, of which the base is the circle ABCD : but
k is also less, for the pyramid is contained within the cone ; which
is impossibla Nor can the cylu^der be less than the triple of the
cone. Let it be less, if possible : therefore, inversely, the cone is
greater than the third part of the cylinder. In the circle ABCD de-
scribe a square ; this square is greater than the half of the circle : and
upon the square ABCD erect a pyramid having the same vertex with
the cone : this pyramid is greater than the half of the cone ; because, as
was before demonstrated, if a square be described about the circle,
the square ABCD is the half of it ; and if, ^ H

upon these squares there be erected solid
paralldojApeds of the same altitudes with
the cone, which are also prisms, the prism
upon the square ABCD shall be the half
of that which is upon the square described E}
about the circle ; for they are to one another
as their bases (33. 11.); as are also the
third parts of them ; thereforethe pyramid,
the base of which is the square ABCD, is
half of the pyramid upon the square de- *
scnl)pd about the circle: but this last pyra-
mid is greater than the cone which it contains ; therefore the p3rra-
mid upon the square ABCD, having the same vertex with the cone,
is greater than the half of the cone. Bisect the circumferenceis AB,
BC, CD, DA in the points E, F, G, H, and join AE, EB, BF, FC, CG,
GD, DH, HA : therefore each of the triangles AEB, BFC, CGD, DHA
is greater than half of the segment of the circle in which it is : upon
each of these triangles erect pyramids having the satae vertex with
the cone. Therefore each of these pyramids is greater than the half
of the segment of the cone in which it is, as before was demonstrated
of the prisms and segments of the cylinder : and thus dividing each
of the circumferences into two equal parts, and joining the points of
division and their extremities by straight lines, and upon the trian^es
erecting pyramids having their vertices the same with that of the
cone, and so on, there must at length remain some segments of the

9ie

THE BLEMfiNTB OF BUOLID.

BOOK XII*

ccffief which together shall be less than the excess of the cone above
the third part of the cylinder. Let these be the s^ments upon AE»
EB, BF, FC, CG, GD, DH, HA. Therefore the rest of the cone, that
is, the pyramid, of which the base is the
polygon AEBFCGDH, and of which the
vertex is the same with that of the cone,
is greater than the third part of the cylin-
der. But this pyramid is the third part of
the prism upon the same base AEBFCGDH, E
and of the same altitude with the cylinder.
Therefore this prism is greater than the
cylinder of which the base is the circle
ABCD. But it is also less ; for it is con-
tained within the cylinder ; which is im-
possible. Therefore i^e cylinder is not less than the triple of the
cone. And it has been demonstrated that neither is it greater than
the triple. Therefore the cylinder is triple of the cone, or the cone
is the third part of the cylinder. Wherefore every cone, &c. Q. £. D.

"==^

,^c|

PROP. XI. THEOR.

CoHTES and cylinders of the same altitude, are to one another
as their bases.*

Let the cones and cylinders, of which the bases are the circlesr
ABCD, EFGH, and the axes KL, MN, and AC, EG the diameters of
their bases, be of the same altitude. As the circle ABCD to the cir-
cle EFGH, so is the cone AL to the cone EN.

If it be not *), let the circle ABCD be to the circle EFGH, as the
cone AL to some solid either less than the cone EN, or greater than
it. First, let it be to a solid less than EN, viz. to the solid X ; and
let Z be the solid which is equal to the excess of the cone EN above
the solid X; therefore the cone EN is equal to the solids X, Z to-
gether. In the circle EFGH describe the square EFGH, therefore
this square is greater than the half of the circle : upon the square
EFGH erect a pyramid of the same altitude with the cone ; this pyra^
mid is greater than half of the cone. For, if a square be described
about the circle, and a pyramid be erected upon it, having the same
vertex with the cone,t the pyramid inscribed in the cone is half of the
pyramid curcumscribed about it, because they are to one another as
their bases (6. 12.) : but the cone is less than the circumscribed pyra-
mid ; therefore the pyramid of which the base is the square EFGH,
and its vertex the same with that of the cone, is greater than half of
the cone: divide the circumferences EF,. FG, GH, HE, each into twa

* See Note.

t Vertex is put in the place of altitude, which is in the Greek, because the pyramidv
in what follows, is supposed to be circumscribed about the cone, and so must hav«
the same vertex. And the same change is made in some places fi>Uowiiig.

BOOK Xil.

TIU 1UIWBIT8 or BOOUD.

tlT

eqnal pafte in th* pdnts O, P, R, B, and join BO, OP, FT, PO, OR*
RH, HS, SE: tb^refbre each of the trIaiigieB BOF, FPG» QRH, H8B
is groi^r than half of th« segmant of the ditle in \i4iich it ta : ^f(m

^ E

each of these triangles erect a p3mimid having tMfe same vertex with
the cone ; each of these pyramids is greater than the half of the seg-
ment of the cone in which it is : and thus dividing each of these cir-
comferences into two equal parts, and from the points of division
drawing straight lines -to the extrenAies of the circumferences, and
upon each of the triangles thus made erecting pyramids, having the
same vertex with the cone, and so on, there must at length remain
sooMTsegments of the cone which are together less (Lem. 1.) than
the solid Z : let these be the segments upon EG, OF, FP, PG, GR,
RH, HS, SB: therefore the remainder of the cone, viar. the pyramid
of which the base is the polygon EOFPORHS, and its vertex the
same with that of the cone, is greater than the solid X : in the circle
ABCD describe the polygon ATBYCVDa similar to the polygon
EOFPGRHS, and upon it erect a pyramid having the same vertex
AL : and because as the square of AC is to the square of EG, so (I.
12.) is the polygon ATBYCVDQ to the polygon BOFPGHHS ; and
as tlfe square of AC to the square of EG, so is (2. 12.) the circle
ABCD to the circle EPGH ; thereibre the circle ABCD (11. 6.) is to
the cffde EFGH, as the polygon ATBYCVDCl to the polygon
EOFPGRHS : but as the circle ABCD to the cirde EFGH, so is the
cone AL to the solid X, and as the polygon ATBYCVDd to the
polygon EOFPGRHS, so is (6. 12.) the pyramid of which the base
is the first of these polygons, and vertex L, to the pyramid of which

ai8

THB ELBHSMTB OP CaCLID.

B0<9K XII.

thebaselsthepther polygon, and its vertex. N: therefore, as the
cone AL to the solid X, so is the pyramid of which the base is the
polygon ATBYCVDQ, and vertex L, to the pyramid the base of
which is the polygon EOFPGRHS, and vertex N : but the cone AL
is greater than the pyramid contauied in it; therefore the solid X is

N] ''^m

greater (14. 5.) than the pyramid In the cone EN ; but it is less, as
was shown, which is absurd : therefore the circle ABCD is not to
the circle EFGH, as the cone M^ to any solid which is less than the
cone EN. In the same manner it may be demonstrated that the
circle EFGH is not to the circle ABCD, as the cone EN to any solid
less than the cone AL. Nor can the circle ABCD be to the circle
EFGH, as the cone AL to any solid greater than the cone EN : for,
if it be possible, let it be so to the solid I, which is greater than the
cone EN : therefore, by inversion, as the circle EFGH to the circle
ABCD, so is the solid I to the cone AL : but as the solid I to the
cone AL, so is the cone EN to some solid which must be less (14.
5.) than the cone AL, because the solid I is greater than the cone
EN : therefore as the circle EFGH is to the circle ABCD, so is the
cone EN to a solid less than the cone AL, which was shown to be
impossible ; therefore the circle ABCD is not to the circle EjjPXxH,
as the cone AL is to any solid greater than the cone EN : and it
has been demonstrated that neither is the circle ABCD to the cir-
cle EFGH, as the cone AL to any solid less than the cone EN :
therefore the circle ABCD is to the circle EFGH, as the cone
AL to the cone EN : but as the cone is to the cone, so (15. 6.) is
the cylmder to the cylinder, because the cylinders are triple (10.

BOOK XII.

THE ELBMBIVTS Ot* EOCLID.

SiO

12.) <^the oonei eadh to each. Therefore, as the circle ABOD to the
circle EFGH; so are the cylinders upon them of the same altitude.
Wherefore cones and cylinders of the same altitude are to one
another as their bases. Q. E. D.

PROP. XII. THEOR.

Similar cones and cylinders have to one another the triplicate
ratio of that which the diameters of their bases have.*

Let the cones and cylinders of which the bases are the circles
ABCD, EFGH, and the diameters of the bases, AC, EG, and KL,
MN the axes of the cones or cylinders, be similar : the cone of which
the base is the circle ABCD, and vertex the point L, has to the cone
of which the base is the circle EFGH, and vertex N, the triplicate
ratio of that which AC has to EG.

For, if the cone ABCDL has not to the cone EFGHN the triplicate
ratio of that which AC has to EG, the cone ABCDL shall have the
triplicate of that ratio to some solid which is less or greater than the
ccme EFGHN. First, let it have it to a less, viz. to the solid X
Make the same construction as in the preceding proportion, and it
may be demonstrated the very same way as in that proposition, that
the pyramid of which the base is the polygon EOFPGRHS, and ver^

rczii

* See Note.

tM Tn fivtxijwmi of e«oud. boob xu*

lex N, te gimler thou the soKd X. Deseribe also in tho dvcb ABCD
the polygon ATBYCVDa «unyar to tbe poAygon EC^PGRHS, opoft
whiob erect a pyramid having the vame vertex with the eone ; and
let LAQ be one of the triangles, containing the pyramid upon tiie
polygon ATBYC VDQ, the vertex of which is L ; and let NES be one
of the triangles containing the pyramid upon the polygon EOFPGRHS
of which the vertex is N ; and join KQ, MS : because then the cone
ABCDL is simflar to the cone EFGHN, AC is (24. deC U.) to EG,
as the axis KL to the axis MN; and as AC to EG, so (15. 5,) is AK
to EM; therefore as AE to EM, so is EL to MN; and, alternately,
AR to KL, as EM to MN : and the right angles AKL, EMN are equal ;
therefore the sides about these equal angles being proportionals, the
triangle AKL is similar (6. 0.) to the triangle EMN. Again, because
AK is to KQ, as EM to MS, and that these sides are about equal
aagles AKQ, EMS, because these angles are, each of them, the same
part of four right angles at the centres K, M ; therefore the triangle
AKCl is similar (6. 6.) to the triangle EMS : and because it has been
shown, that as AK to KL, so is EM to MN, and that AK is equal to
SO, and EM lo MS, as QK to KL, so is SM to MN, and therefore
the sides about the right angles QKL, SMN being propoitioneds, the
triangle LKQ is similar to the triangle NMS : and beeause of the
tftmilartty of the triangles AKL, EMN, as LA is to AK, so is NB to
BM : and by the similarity of the triangles AKQ, EMS, as KA to ACl,
ao ME to ES ; ex saquali, (22. 5.) LA is to AQ as NE to ES. Again,
because of the similarity of the triangles L€tK, NSM; as LQ to QK,
so NS to SM ; and from the sunilarity of the triangles KAQ, MES,
as KCt to QA, so MS to SE; ex mqwdi^ (22. 5.) LQ. is to QA, as
NS to SE : and it was proved that QA is to AL as SE to EN, there-
fore, again, ex mquali, as QL to LA, so is SN to NE ; wherefore the
triangles LQA, NSE, Having the sides about all their angles propor-
tionals, are equiangular (5. 6.) and similar to pne another: and there-
fore the pyramid of which the base is the triangle AKQ and vertex
L, is similar to the pyramid the base of which is the triangle EMS,
and vertex N, because their solid angles are equal (B. 11.) to one
another, and they are contained by the same number of similar
planes : but similar pyramids which have triangular bases have to
one another the triplicate (8. 12.) ratio of that which their homologous
sides have ; therefore the pyramid AKQL has to the pyramid EMSN
tlie triplicate ratio of that which AK has to EM. In the same man-
ner, if straight lines be drawn from the points D, V, C, Y, B, T to K,
and from the points H, R, G, P, F, O to M, and p3rramids be erected
upon the triangles having the same vertices with the cones, it may
be demonstrated that each pyramid in the first cone has to each in
the other, taking them in the same order, the triplicate ratio of that
which the side AK has to the side EM ; that is, which AC has to EG:
but as one antecedent to its consequent, so are all the antecedents
to all the consequents fl2. 5.); therefore as the pjottmid AKQL to
the pyramid EMSN, so is the whple pyramid the base of which
is the polygon DQATBYCV, and vertex L, to the whole pyramid

BOOK* ma.

THB EUlMBNTt OP BUCLID.

921

cxf which the base it the pdygon HSB0FP6R, and vertex N. Where-
fore also the first of these two last named pyramids has to the other
the triplicate ratio of that which AC has to EG. But by the hypo-
thesis, the cone of which the base is the circle ABCD, and vertex L, has
to the sdid X, the triplicate ratio of that which AC has to £SG : there*
fore as the cone of which the base is the circle ABCD, and vertex
L, is to the solid X, so.is the pyrsunid the base of which is the poly*
gon DQATBYCV, and vertex L, to the pyramid the base of which
is the polygon HSEOFPGR, and vertex N : but the said cone is
greater than the pyramid contained in it, therefore the solid Z is
gfoater (14. 5.) than the pyramid, the base of which is the polygon
HSBOFPOR, and vertex N ; but it is also less; which is impossible;
therefore the cone, of which the base is the cirde ABCD, and vertex
U ha3 not io any solid which is less than the cone of which the base

is the circle EFGH and vertex N, the triplicate ratio of that which
A€ has to £Q. In the same manner it may be demonstrated that
neither has the cone EFGHN to any solid which is less than the
cone ABCDLi, the triplicate ratio of that which EG has to AC. Nor
can the cone ABCDL have to any solid which is greater than the
cone EFGHPT, the triplicate ratio of that which AC has to EG : for,
tf it be possible, let ft have it to a greater, viz. to the solid Z : there-
fore, inversely, the solid Z has to the cone ABCDli, the triidicate
ratio of that which EG has to AC : but as the solid Z is to the cone

222

THE ELEMENTS OF EUCLID.

BOOK XII.

ABCDL, SO is the cone EFGHN to some iiolld, which must be less
(14. 5.) than the cone ABCDL, because the solid Z is greater than
the cone EFGHN : therefore the cone EFGHN has to a solid which
is less than the cone ABCDL, the triplicate ratio of that which EG
has to AC, which was demonstrated to be impossible : therefore the
cone ABCDL has not to any solid greater than the cone EFGHN,
the triplicate ratio of that which AC has to EG ; and it was demon-
strated that it could not have that ratio to any solid less than the
cone EFGHN: therefore the cone ABCDL has to the cone EFGHN
the triplicate ratio of that which AC has to EG : but as the cone is
to the cone, so (15. 5.) the cylinder to the cylinder; for every cone
is the third part of the cylinder upon the same base, and of the same
altitude : therefore also the cylinder has to the cylinder the triplicate
ratio of that which AC has to EG. Wherefore similar cones, &c.
Q.. £. D.

PROP. Xm. THEOR.

If a cylinder be cut by a plane, parallel to its opposite planes,
or bases, it divides the cylinder into two cylinders, one of which
is to the other as the axis of the first to the axis of the other.*

Let the 9ylinder AD be cut by the plane
GH, parallel* to the opposite planes AB, CD, O
meeting the axis EF in the point K, and let the
line GH be the common section of the plane
GH and the surface of the cylinder AD : let
AEFC be the parallelogram, in any position of R
it, by the revolution of which about the straight
line EF the cylinder AD is described ; and let
GK be the common section of the plane GH,
and the plane AEFC : and because the paral- A
lel planes AB, GH are cut by the plane AEKG,
AE, KG, their common sections with it, are
parallel (16. 11.): wherefore AK is a paral-
lelogram, and GE equal to ElA. the straight G
line from the centre of the circle AB : for the
same reason each of the straight lines drawn q
from the point K to the line GH may be proved
to be equal to those which are drawn firom the
centre of the circle AB to its circumference, ^
and are therefore . all equal to one another.
Therefore the line GH is the circumference of ^
a circle, (15. def. 1.) of which the centre is the point K : therefore
the plane GH divides the cylinder AD into the cylinders AH, GD ;
for they are the same which would be described by the revolution
of the parallelograms AK, GF, about the straight lines EK, KP: and

* See Note.

BOOK XII.

THE ELEM£J<TS OP EUCLID.

223

it is to be shown, that the* cylinder AH is to the cylinder HC, as
the axis EK to the axis KF.

Produce the axis EF both ways ; and take any number of straight
lines EN, NL, each equal to EK; and any number FX, XM each
equal to FK; and let planes parallel to AB,
CD pass through the points L, N, X, M ; there- O
fore the common sections of these planes with
the cylinder produced are circles the centres
of which are the points L, N, X, M, as was R
proved of the plane GH : and these planes cut
off the cylinders PR, RB, DT, TQ : and be-
cause the axes LN, NE, EE are all equal, there-
fore the cylinders PR, RB, BG are (11. 12.) to A
one another as their bases : but their bases are
equal, and therefore the cylinders PR, RB, BG
are equal: and because the axes LN, NE, EK
are equal to one another, as also the cylinders
PR, RP, BG, and that there are as many axes
as cylinders ; therefore, whatever multiple the p
axis KL is of the axis KE, the same multiple
is the cylinder PG of the cylinder GB : for the
same reason, whatever multiple the axis MK T
is of tjhe axis KF, the same multiple is the cy-
linder Q.G of the cylinder GD: and if the axis V
KL be equal to the axis KM, the cylinder PG
is equal to the cylinder GQ ; and if the axis KL Be greater than the
axis KM, the cylinder PG is greater than the cylinder QG ; and if less^
less : since therefore there are four magnitudes, viz. the axes EK, KF,
and the cylinders BG, GD, and that of the axis EK and cylinder BG,.
there has been taken any equimultiples whatever, viz. the axis KL
and cylinder PG ; and of the axis KF and cylinder GD, any equimul-
tiples whatever, viz. the axis KM and cylinder GQ : and it has been
demonstrated, ijf the axis KL be greater than the axis KM, the cylin-
der PG is greater than the cylinder GQ;^and if equal, equal; and if
less, less : therefore (5. def. 5.) the axis EK is to the axis FK, as the
cylinder BG to the cylinder GD. Wherefore, if a cylinder, &c^
€t KD.

k«

PROP. XIV. THEOR.

Cones and cylinders upon equal bases are to one another as
their altitudes.

Let the cylinders EB, FD be upon the equal bases AB, CD : as the
cylinder EB to the cylinder FD, so is the axis GH to the axis KL.

Produce the axis KL to the point N and make LN equal to the
axis GH, and let CM be a cylinder of which the base is CD, and axis.
LN : and because the cylinders EB, CM,, have the same altitude, they
are to one another as their bases : (11. 12.) but their bases are^qual :

224

THE ELBMENT8 OF EUCLID.

BOOK XIL

therefore also the cylinders £B, CM are equal. And becauae the
cylinder PM is cut by the plarfe CD
parallel to its opposite planes, as the
cylinder CM to the cylinder PD, so
is (13; 12.) the axis LN to the axis
KL. But the cylinder CM is equal
to the cylinder £B, and the axis LN E
to the axis GH : therefore as the cy-
linder EB to the cylinder PD, so is
the axis GH to the axis KL : and as
the cylinder EB to the cylinder FD^
so is (16. 5.) the cone ABG to the
cone CDK, because the cylinders
are triple (10. 12.) of the cones: A
therefore also the axis GH is to the
axis EL, as the cone ABG to the cone CDK, and the cylinder EB to
the cylhider PD. Wherefore cones, &c. d. E. D.

PROP. XV. THEOR.

The bases and altitudes of equal cones and cylinders are re-
ciprocally proportional; and, if the bases and altitudes be recipro-
cally proportional, the Qones and cylinders are equal to one
another.*

Let the circles ABCD, EFQH, the diameters of which are AC, EG,.
be the bases, and KL, MN the axes, as also the altitudes of equal
cones and cylinders ; and let ALC, ENG be the cones, and AX, EG
the cylinders : the bases and altitudes of the cylinders AX, EG are
reciprocally proportional; that is, as the base ABCD to ihe base
SPGH, so is the altitude MN to the altitude KL.

Either the altitude MN is equal to the altitude KL, or these altitudes
are not equal. Pirst, let them be equal : and the cylinders AX, E0»
being also equal, and cones and cylinders of the same altitude being
to one another as their bases,
(1 1. 12.) therefore the base ABCD
is equal (A. 5.) to the base
EFGH ; and as the base ABCD
is to the base EPGH, so. is the
altitude MN to the altitude KL.
But let the altitudes KL, MN
be unequal* and MN the greater
of the two, and from MN take
MP, equal to KL, and, through
the point P, cut the cylinder EO A^
by the plane TYS, parallel to the
opposite planes of the circles
EPGH, RO; therefore the com-
mon section of the plane TYS and the cylinder EO is a drete, and

* See Note.

HOOK Xa, TBB ELBMBim OF HOOLID.

consequently ES is a cylinder, the base of whkh is the drcle EFHQ,
and altitude MP : and because the cylinder AX is equal to the cylin-
der £0, as AX is to the cylinder ES, so (7. 5.) is the cylinder £0 to
the same ES. But as the cylkider AX to the cylinder ES, so (II.
12.) is the base ABCD to the base EFGH ; for the cylinders AX, ES
are of the same altitude ; and as the cylinder EO to the cylinder ES,
so (13. 12.) is the altitude MF to the altitude MP, because the cyUn-
der EO is cttt by the plane TYS parallel to its opposite i^nes.
Therefore as the base ABCD to the base EFGH, so is the altitude
MN to the altitude MP : but MP is equal to the altitude KL : where-
fore as the base ABCD to the base EFGH, so is the altitude MN to
the altittide KL : that is, the bases and altitudes of the equal cyiin^
ders AX, EO are reciprocally proportional.

But let the bases and altitudes of the cylinders AX, EO be recipro-
cally proportional, viz. the base ABCD to the base EFGH, as the al-
titude MN to the altitude KL : the cylinder AX is equal to the cylin-
der EO.

First, let the base ABCD be equal to the base EFGH ; then be-
cause as the base ABCD is to the base EFGH, so is the altitude MN
to the altitude KL ; MN is equal (A. 5.) to KL, and therefore the
cylinder AX is equal (11. 12.) to the cylinder EO.

But let the bases ABCD, EFGH be unequal, and let ABCD be the
greater ; and because as ABCD is to the base EFGH, so is the altitude
MN to the altitude KL ; therefore MN is greater (A. 5.) than KL.
Then, the same construction being made as before, because as the
base ABCD to the base EFGH, so is the altitude MN to the altitude
KL ; and because the altitude KL is equal to the altitude MP ; there-
fore the base ABCD is (11. 12.) to the base FiFGH, as the cylinder
AX to the cylinder ES ; and as the altitude MN to the altitude MP
or KL, so is the cylinder EO to the cylinder ES : therefore tlie cylinr
der AX is to the cylinder ES, as the cylinder EO is to the same ES;
whence the cylinder AX is equal to the cylinder EO ; and the same
reasoning holds in cones. Q. E. D.

PROP. XVI. PROB.

To describe in the greater of the two circles that have the
same centre, a polygon of an even number of equal sides, that
shall not meet the lesser circle.

Let ABCD, EFGH be two given circles having the same centre
K : it is required to inscribe in the greater circle ABCD a polygon
of an even number of equal sides, that shall not meet the lesser circle.

Through the centre K draw the straight hue BD, and from the
point G, where it meets the circumference of the lesser circle, draw
GA at right angles to BD, and produce it to C ; therefore AC touches
(16. 3.) the circle EFGH: then, if the circumference BAD be bisect-
ed, and the half of it be again bisected, and so on, there must at
length remain a circumference less (Lemma.) than AD : let this be

THE BLBMENTS OF SUbLID. BOOK XII.

LD; and from ttie point LdrawLM per-
pendicular to BD» and produce it to N ;
and join LD, DN. Therefore LD is
equal to DN ; and because LN is paral-
lel to AC, and that AC touches the circle BL
EFGH, therefore LN does not meet the
circle EFQH : and much less shall the
straight lines LD, DN meet the circle
£FQH, so that if straight lines equal to
LD be applied in the circle ABCD from
the point L around to N, there shall be described in the circle a poly-
gon of an even number of equal sides not meeting the lesser circle.
Which was to be done.

LEMMA IL

If two trapeziums ABCD, EFGH be inscribed in the circles,
the centres of which are the points K, L ; and if the sides AB,
DC be parallel, as also EP, HG : and the other four sides AD,
BC, EH, F6 be all equal to one another ; but the side AB greater
than EP, and EC greater than HG ; the straight line KA from
the centre of the circle in which the greater sides are, is greater
than the straight line LE drawn from the centre to the circum-
ference of the other circle.

If it be possible, let EA be not greater than LE ; then KA must be
either equal to it or less. First, let EA be equal to LE : therefore
because in two equal circles, AD, DC in the one, are equal to EH,
FG in the other, the circumferences AD, BC are equal (28. 3.) to the
circumferences EH., FG ; but because the straight lines AB, DC are
respectively greater than EF, GH, the circumferences AB, DC are
greater than EF, HG : therefore the whole circumference ABCD
is greater than the whole EFGH ; but it is also equal to it, which
is impossible : therefore the straight line EA is not equal to LE.

But let EA be less than LE, and make LM equal to EA, and from
the centre L, and distance LM, describe the circle MNOP, meeting
the straight lines LE, LF, LG, LH, in M, N, O, P ; and join MN,
NO, OP, PM, which are respectively parallel (2. 6.) to and less than
EF, FG, GH, HE : then because EH is greater than MP, AD is
greater than MP; and the circles ABCD, MNOP are equal: there-
fore the circumference AD is greater than MP ; for the same reason,
the circumference BC is greater than NO ; and because the straight
line AB is greater than EF, which is greater than MN, much more is
AB greater than MN : therefore the circumference AB is greater than
MN ; and«for the same reason, the circumference DC is greater than
PO : therefore the whole circumference ABCD is greater than the

BOOK XII.

THE ELEBI KNTS QF EUCLID.

whole MNOP; but it Is likewise equal to it, whicli is impossible:
therefore KA is not less than LB ; nor is it equal to it : the straight
line KA must therefore be greater than LE. Q. E. D.

Cor. And if there be an isosceles triangle, the sides of which are
equal to AD, BC, but its base less than AB the greater of the two
sides AB, DC ; the straight line KA may, in the same manner, be
demonstrated to be greater than the straight line drawn from the
centre to the circumference of the circle described about the triangle.

PROP. xvn. PRoa

To describe in the greater of two spheres which have the
some centre, a solid polyhedron, the superficies of which shall
not meet the lesser sphere.*

Let there be two spheres about the same centre A ; it is required to
describe in the greater a solid polyhedron, the superficies of which
shall not meet the lesser sphere.

Let the spheres be cut by a plane passing through the centre ; the
common sections of it with the spheres shall be circles : because the
sphere is described by the revolution of a semicircle about the diame-
ter remaining unmoveable : so that in whatever position the semi-
circle be conceived, the common section of the plane in which it is
with the superficies of the sphere is the circumference of a circle ;•
and this is a great circle of the sphere, because the diameter of the
sphere, which is likewise the diameter of the circle, is greater (15.
3.) than any straight line in the circle or sphere : let then the circle
made by the section of the plane with the greater sphere be BCDE,
and with the lesser sphere be FGH ; and draw the two diameters

BD, CE at right angles to one another; and in BCDE, the greater of
the two circles, describe (16. 12.) a polygon of an even number of
equal sides, not meeting the lesser circle FGH ; and let its sides, in

BE, the fourth part of the circle, be BK, KL, LM, ME ; joui KA and
produce it to N ; and from A draw AX at right angles to the plane
of the curcle BCDE, meeting the superficies of the sphere in the point

* See Note.

Tfld ELBMEN19 OP BOCLID.

BOOK ni.

X; and let tiie planes pass through AX, and each of the straight lines
BD» KN, which, from wh^t has been said, shall produce great circles
on the superficies of the sphere; and let BXD, KXN be the semicir-
cles thus made upon the diameters BD, KN : therefore, because XA
is at right angles to the plane of the circle BCDE, every plane which
passes through XA is at right (18. 11.) angles to the plane of the
circle BODE ; wherefore the semicircles BXD, KXN, are at right an-
gles to that plane ; and because the semicircles BED, BXD, KXN,
upon the equal diameters BD, KN are equal to one another, their
halves BE, BX, KX, are equal to one another ; therefore, as many
sides of the polygon as are in BE, so many there are in BX, KX
equal to the sides BK, KL, LM, ME: let these polygons be described,
and their sides be BO, OP, PR, RX ; KS, ST, T Y, YX, and join OS,
PT, RY ; and from the points O, S, draw OV, SO, perpendiculars to
AB, AK: and because the plane BOXD is at right angles to the
plane BCDE, and in one of them BOXD, OV is drawn perpendicular
to AB the common section of the planes, therefore OV is perpendicu-
lar (4. def. 11.) to the plane BCDE: for the same reason SQ is per-
pendicular to the same plane, because the plane KSXN is at right
angles to the plane BCDE. Join VQ; and because in the equal semi-
circles BXD, KXN the circumferences BO, KS are equal, and OV,

set are perpendicular to their diameters, therefore (26. L) OV is
equal to SQ, and B V equal to KQ. : but the whole BA is equal to tiie
whole KA, therefore the remainder VA is equal to the remainder

BOOK ZII*

THfl BLEMBNTS OF BVCLID.

QA ; as ther^ore BY is to VA, so is KQ> to QA, wherefore VQ is
parallei (2. 6.) to BK ; and because O V, SQ are each of them at
right angles Xo the plane of the circle BCDE, OV is parallel (6. 11.)
to SO, ; and it has been proved that it is also equal to it ; therefore
Q^Vj SO are equal and parallel (33. I.) : and because QV is parallel
to SO, and ieaso to KB, OS is parallel (9. 11.) to BK ; and therefore
BO, KS which join them are in the same plane in which these paral-
lels are, and the quadrilateral figure KBOS is in one plane; and if
PB, TK be joined, and perpendiculars be drawn from the points P,
T to the straight lines AB, AK, it may be demonstrated, that TP is
parallel to KB in the very same way that SO was shown to be pa-
rallel to the same KB ; wherefore (9. 11.) TP is parallel to SO, and
the quadrilateral figure SOPT is in one plane : for the same reason,
the quadrilateral TPRY is in one plane ; and the figure YRX is also
in one plane (2. 11.). Therefor^ if from the points O, S, P, T, R, Y

there be drawn straight lines to the point A, there shall be formed a
solid polyhedron between the curcumferences BX, KX composed of
pyramids, the bases of which are the quadrilaterals KBOS, SOPT,
TPRY, and the triangle YRX, and of which the common vertex is
the point A : and if the same construction be made upon each of the
sides KL, LM, ME, as has been done upon BK, and the like be done
also in the other three quadrants, and in the other hemisphere ; there
shall be formed a solid polyhedron described in the sphere,^ompo8ed
of pyramids, the bases of whidi are the aforesaid quadrilateral figures

930 THE BLBMENTS OF EUCLID. BOOK ZII.

and the triangle YRX, and those formed in the like manner in the
rest of the sphere, the common vertex of them all being the point A;
and the superficies of this solid polyhedron does not meet the lesser
sphere in which is the circle FGH ; for, from the point A draw (11.
11.) AZ perpendicular to the plane of the quadrilateral KBOS» meet-
ing it in Z, and join BZ, ZK : and because AZ is perpendicular to
the plane KBOS, it makes right angles with every straight line meet-
ing it in that plane ; therefore AZ is perpendicular to BZ and ZK ;
and because AB is equal to AK, and that the squares of AZ, ZB are
equal to the square of AB ; and the squares of AZ, ZK to the square
of AK (47. 1.): therefore the squares of AZ, ZB are equal to the
squares of AZ, ZK : take from these equals the square of AZ, the
remaining squeire of BZ is equal to the remaining square of ZK ;
and therefore the straight line BZ is equal to ZK : in the like man-
ner it may be demonstrated, that the straight lines drawn from the
point Z to the points O, S, are equal to BZ or ZK : therefore the
circle' described from the centre Z, and distance ZB, shall pass
through the points K, O, S, and KBOS shall be a quadrilateral figure
in the circle : and because KB is greater than QV, and ^dV equal
to SO, therefore KB is greater than SO ; but KB is equal to each of
the straight lines BO, KS ; wherefore each of the circumferences cut
off by KB, BO, KS is greater than that cut off by OS ; and these
three circumferences, together with a fourth equal to one of them,
are greater than the same three together with that cut off by OS ;
that is, than the whole circumference of the circle ; therefore the
circumference subtended by KB is greater than the fourth part
of the whole circumference of the circle KBOS, and consequently
the angle BZK at the centre is greater than a right angle: and
because the angle BZK is obtuse, the square of BK is greater (12.
2.) than the squares of BZ, ZK ; that is, greater than twice the
square of BZ. Join KV, and because in the triangles KBV, OBV,
KB, BV are equal to OB, BV, and that they contain equal angles ;
the angle KVB is equal (4. 1 .) to the angle OVB : and OVB is a
right angle ; therefore also KVB is a right angle : and because BD
is less than twice DV, the rectangle contained by DB, BV is less
than twice the rectangle DVB ; that is (8. 6.), the square of KB is
less than twice the square of KV : but the square of KB is greater
than twice the square of BZ ; therefore the square of KV is greater
than the square of BZ : and because BA is equal to AK, and that
the squares of BZ, ZA are equal together to the square of BA, and
the squares of KV, VA to the square of AK : therefore .the squares
of BZ, ZA are equal to the squares of KV, VA ; and of these the
square of KV is greater than the square of BZ ; therefore the square
of VA is less than the square of Z A, and the straight line AZ great-
er than VA : much more th^n is AC greater than AG ; because, in
the preceding proposition, it was shown that KV falls without the
circle FGH : and AZ is perpendicular to the plane KBOS, and is
therefore the shortest of all the straight lines that can be drawn from
A, the centre of the sphere, to that plane. Therefore the plane KBOS
does not meet the lesser sphere.

BOOK XII. THE ELEMENTS OF EUCLID. 231

And that the other planes between the quadrants BX, KX fall with-
out the lesser sphere, is thus demonstrated ; from the point A draw
AI perpendicular to the plane of the quadrilateral SOFT, and join
10 ; and, as was demonstrated of the plane KBOS, and the point Z,
in the same way it may be shown that the point I is the centre of a
circle described about SOFT ; and that OS is greater than FT ; and
FT was shown to be parallel to OS ; therefore, because the two trape*
ziums KBOS, SOFT inscribed in circles have their sides BK, OS, pa^
rallel, as also OS, FT ; and their other sides BO, KS, OF, ST aU equal
to one another, and that BK is greater than OS, and OS greater than
FT, therefore the straight line ZB is greater (2. lem. 12.) than lO.
Join AO which will be equal to AB : and because AIO, AZB are right
angles, the squares of Al, 10 are equal to the square of AO or of AB;
that is, to the squares of AZ, ZB: and the square of ZB is greater
than the square of 10, therefore, the square of AZ is less than the
square of AI; and the straight line AZ less than the straight line AI;
and it was proved that AZ is greater than AG : much more then is
AI greater than AG : therefore the plane SOFT falls wholly without
the lesser sphere r in the same manner it may be demonstrated that
the plane TFRY falls without the same sphere, as also the triangle
YRX, viz. by the cor. of 2d lemma. And after the same way it may
be demonstrated that all the planes which contain the solid polyhe-
dron, fall without the lesser sphere. Therefore in the greater of the
two spheres which have the same centre, a solid polyhedron is de^
scribed, the superficies of which does not meet the lesser sphere.
Which was to be done.

But the straight line AZ may be demonstrated to be greater than
AG, otherwise, and in a shorter manner, without the help of prop.
16, as follows. From the point G draw GU at right angles to AG,
and join AU. If then the circumference BE be bisected, and its half
again bisected, and so on, there will at length be left a circumference
less than the circumference which is subtended by a straight line
equal to GU, inscribed in the circle BCDE: let this be the circumfe-
rence KB ; therefore the straight luie KB is less than GU ; and be-
cause the angle BZK is obtuse, as was proved in the preceding,
therefore BK is greater than BZ : but GU is greater than BK ; much
more then is GU greater than BZ, and the square of GU than the
square of BZ, and AU is equal to AB; therefore the square of AU,
that is, the squares of AG, GU are eqvial to the square of AB, that is,
to the squares of AZ, ZB ; but the square of BZ is less than the
square of GU ; therefore the square of AZ is greater than the square
of AG, and the straight line AZ consequently greater than the straight
line AG.

CoK. And if in the lesser sphere there be described a solid poly-
hedron, by drawing straight lines betwixt the points in which the
straight lines from the centre of the sphere drawn to all the angles
of the solid polyhedron in the greater sphere meet the superficies of
the lesser ; in the same order in which are joined the points in which
the same lines froln the centre meet the superficies of the greater
sphere; the solid polyhedron in the sphere BCDE has to this other
solid polyhedron the triplicate ratio of that which the diameter of the

THB XLEMSim OF SDCLID. BOOK XII.

sphere BCDE has to the diameter of the other sfihere; for if these
two solids be divided into the same number of pjrramids, and in the
same order, the pyramids shall be similar to one another each to
each ; because they have the solid angles at their common vertex,
the centre of the sphere the same in each pyramid, and their other
solid angle at the bases equal to one another, each to each (B. 11.),
because they are contained by three plane angles equal each to
each : and the pyramids are contained by the same number of similar
planes ; and are therefore similar (11. def. 11.) to one another, each
to each : but similar pyramids have to one another the triplicate (Cor.
8. 12.) ratio of their homologous sides. Therefore the pyramid of
which the base is the quadrilateral KBOS, and vertex A, has to the
pyramid in the other sphere of the same order, the triplicate ratio of
their homologous sides, that is of that ratio which AB from the centre
of the greater sphere has to the straight line from the same centre to
the superficies of the lesser sphere. And in like manner, each pyra-
mid in the greater sphere has to each of the same order in the lesser,
the triplicate ratio of that which AB has to the semi-diameter of the
lesser sphere. And as one antecedent is to its consequent, so are all
the antecedents to all the consequents. Wherefore the whole solid
polyhedron in the greater sphere has to the whole solid polyhedron in
the other, the triplicate ratio of that which AB, the semi-diameter of
the first, has to the semi-diameter of the other ; that is, which the
diameter BD of the greater has to the diameter of the other sphere.

PROP. XVn. THEOR.

SpH&Ri» have to one another the triplicate ratio of that which
their diameters have.

Let ABC, DEF be two spheres, of which the diameters are BC,
EP. The sphere ABC has to the sphere DEF the triplicate ratio of
that which BC has to EF.

For, if it has not, the sphere ABC shall have to a sphere either
less or greater than DEF, the tiiplibate ratio of that which BC
has to EF. First, let it have that ratio to a less, viz. to the sphere
GHK; and let the sphere DEF have the same centre with GHK;
and in the greater sphere DEF describe (17. 12.) a solid polyhe-
A D L

dron, the superficies of which does not meet the lesser sphere
GHK ; and in the sphere ABC describe another similar to that in

BOOK ZII. THB BLEM1NT9 OP EUCLID. 233

the sphere DEF; therefore the sofid polyhedron hi the sphere ABC
has to the solid polyhedron in the sphere DEF, the triplicate ratio
(Cor. 17. 12.) of that which BC has to EF. But the sphere ABC
has to the sphere GHK, the triplicate ratio of that which BC has to
EF : therefore, as the sphere AJBC to the sphere GHK, so is the solid
polyhedron in the sphere ABC to the solid polyhedron in the sphere
DEF : but the sphere ABC is greater than the solid polyhedron in it ;
therefore (14. 5.) also the sphere GHK is greater than the solid poly-
hedron in the sphere DEF ; but it is also less, because it is contained
within it, which is impossible : therefore the sphere ABC has not to
any sphere less than DEF, the triplicate ratio of that which BC has
to EF. In the same manner, it may be demonstrated that the sphere
DEF has not to any sphere less than ABC, the triplicate ratio of that
which EF lias to BC. Nor can the sphere ABC have to any sphere
greater than DEF, the triplicate ratio of that which BC has to EF :
for, if it can, let it have that ratio to a greater sphere LMN : there-
fore, by inversion, the sphere LMN has to the sphere ABC the tripli-
cate ratio of that which the diameter EF has to the diameter BC.
But as the sphere LMN to ABC, so is the sphere DEF to some
sphere, which must be less (14. 5.) than the sphere ABC, because
the sphere LMN is greater than the sphere DEF. Therefore the
sphere DEF has to a sphere less than ABC the triplicate ratio of that
which EF has to BC ; which was shown to be impossible : therefore
the sphere ABC has not to any sphere greater than DEF the tripli-
cate ratio of that which BC has to EF : and it was demonstrated,
that neither has it that ratio to any sphere less than DEF. There-
fore the sphere ABC has to the sphere DEF, the triplicate ratio of
that which BC has to EF. Q. R D.

FINIS.

NOTES,

CRITICAL aND GEOMETRICAL:

CONTAINING

AN ACCOUNT OP THOSE THINGS IN WHICH THIS EDITION

DIPPERS FROM THE GREEK TEXT;

AND

THE REASONS OF THE ALTERATIONS WHICH HAVE BEEN MADE.

AS ALSO

OBSERVATIONS

ON SOME OF THE PROPOSITIONS.

BY ROBERT SIMSON, M. D.

BMiaiTUS PROFESSOR OF MATHEMATICS IM THE rNIVBRSITY OF OLABOOW.

n '

NOTES, &c.

DEFINITION I. BOOK I.

It is necessary to consider a solid, that is, a magnitude which has
length, broadth, and thickness, in order to understand aright the
definitions of a point, line, and superficies; for these all arise from
a solid, and exist in it : the boundary, or boundaries which contain
a solid are called superficies, or the boimdary which is common to
two solids which are contiguous, or which divides one solid into two
contiguous parts, is called a superficies : thus, if BCGF be one of
the boundaries which contain the solid ABCDEFGH, or which is the
common boundary of this solid, and the solid BKLCPNMG, and is
therefore in the one as well as in the other solid called a superficies,
and has no thickness : for, if it have any, this thickness must either
be a part of the thickness of the solid AG, or the solid BM, or a part
of the thickness of each of them. It H . G M

cannot be a part of the thickness of the Al TfT 71

solid BM ; because, if this solid be re- E j^ 1 f I f \
moved from the solid AG, the super- ' ' * ' ' '
ficies BCGF, the boundary of the solid
AG, remains still the same as it was.
Nor can it be a part of the thickness
of the solid AG ; because if this be re-
moved fi-om the solid BM, the superfi-
cies BCGF, the boundary of the solid BM, does nevertheless remain :
therefore the superficies BCGF has no thickness, but only length
and breadth.

The boundary of a superficies is called a line, or a line is the
common boundary of two superficies that are contiguous, or which
divides one superficies into two contiguous parts : thus, if BC be
one of the boundaries which contain the superficies, ABCD, or which
is the common boundary of this superficies, and of the superficies
B^BCL, which is contiguous to it, this boundary BC is called a line,
and has no breadth : for if it have any, this must be part either of
the breadth of the superficies ABCD, or of the superficies KBCL, or
a part of each of them. It is not part of the breadth (rf the super-
ficies KBCL ; for, if these superficies be removed from the super-
ficies ABCD, the line BC, which is the boundary of the superficies
^^^CDy remains the same as it was : nor can the breadth that BC is

238

NOTES.

BOOK I.

supposed to have, be a part of the breadth of the superficies ABCD ;
because, if this be removed from the superficies KBCL, the hne BC,
which is the boundary of the superficies KBCL, does nevertheless
remain : therefore the line BC has no breadth : and because the line
BC is a superficies, and that a superficies has no thickness, as was
shown ; therefore a line has neither breadth nor thickness, but only
length.
The boundary of a line is called a point, or a point is the common

....^

boundary or extremity of two lines
that are contiguous : thus, if B be the
extremity of the line AB, or the com-
mon extremity of the two lines AB, KB,
this extremity is called a point, and has
no length ; for, if it have any, this
length must either be part of the length
of the line AB, or of the line KB. It is
not part of the length of KB : for if the
line KB be removed from AB, the point
B, which is the extremity of the line AB, remains the same as it
was : nor is it part of the length of the line AB ; for, if AB be re-
moved from the line KB, the point B, which is the extremity of the
line KB, does nevertheless remain: therefore the point B has no
length : and because a point is in a line, and a line has neither
breadth nor thickness, therefore a point has no length, breadth, new
thickness. And in this manner the definitions of a point, line, and
superficies, are to be understood.

DEF. Vn. B. I.

Instead of this definition as it is in the Greek copies, a more dis-
tinct one is given frbm a property of a plane superficies, which is
manifestly supposed in the Elements, viz. that a straight line drawn
from any point in a plane to any other in it is wholly in that plane.

DEF. VIU. B. I.

It seems that he who made this definition designed that it should
comprehend not only a plane angle contained by two straight lines,
but likewise the angle which some conceive to be made by a straight
line and a curve, or by two curve lines, which meet one another in
a plane : but, though the meaning of the words s<ir' su^sio^, that is, in
a straight line, or in the same direction, be plain, when two straight
lines are said to be in a straight line, it does not appear what ought
to Be understood by these words, when a straight line and a curve,
or two curve lindls, are said to be in the same direction ; at least it
cannot be explained in this place ; which makes it probable that this
definition, and that of the angle of a segment, and what is said of the
angle of a semicircle, and the angles of segments, in the 16th and
31 6t propositions of book 3, are the editions of some less skilful eidi-
tor ; on whicli account, esj)ecially since they are quite useless, these

BOOK I. . NOTES. ^ 83d^

defitiitions are distinguished from the rest by inverted double com-
mas.

DEP. XVII. B. I.

The words, »* which also divide the drcJe into two equal parts," are
added at the end of this definition in all the copies, but are now left
out as not belonging to the definition, being only a corollary from it
Proclus demonstrates it by conceiving one of the parts into which the
diameter divides the circle, to be applied to the other ; for it is plain
^ they must coincide, else the straight lines from the centre to the cir-
cumference would not be all equal : the same thing is easily deduced
from the 31st prop, of book 3, and the 24th of the same; from the
first of which it follows, that semicircles are similar segments of a
circle : and from the other, that they are equal to one another.

DEF. XXXIII. B. I.

This definition has one condition more than is necessary ; because
every quadrilateral figure which has its opposite sides equal to one
another, has likewise its opposite angles equal, and on the contrary.

Let ABCD be a quadrilateral figure, of which the opposite sides
AB, CD are equal to one another : as also AD A D

and BC : join BD ; the two sides AD, DB are
equal to the two CB, BD, and the base AB is
equal to the base CD, therefore, by prop. 8, of
book 1, the angle ADB is equal to the angle ^
CBD; and, by prop. 4, book 1, the angle BAD
is equal to the angle DCB, and ABD to BDC ; and therefore also the
angle ADC is equal to the angle ABC.

And if the angle BAD be equal to the opposite angle BCD, and
the angle ABC to ADC, the opposite sides are equal ; because, by
prop. 32, book 1, all the angles of the quadrilateral figure ABCD are
together equal to four right angles, and the A D

two angles BAD, ADC, are together equal
to the two angles BCD, ABC : wherefore
BAD, ADC are the half of all the four an-
gles ; that is BAD and ADC are equal to
two right angles: and therefore AB, CD B C

are parallels by prop. 28, B. 1. In the same manner, AD, BC are
parallels : therefore ABCD is a parallelogram, and its opposite sides
are equal, by prop. 34, book 1.

PROP. VII. B. I.

There are two cases of this proposition, one of which is not in the
Greek text, but it is as necessary as the other : and that the case left
out has been formerly in the text, appears plainly from this, that the
second part of prop. 5, which is necessary to the demonstration of
this case, can be of no use at all in the Elements, or any \*fhere else.

340 « NOTES. • BOOK I.

but in thi« demonstration : because the second part of prop. 6, clearly
follows from the first part^ and prop. 13, book 1. This part must
therefore have been added to prop. 5, upon account of some propo-
sition betwixt the 5th and 13th, but none of these stand in* need of it
except the 7th proposition, on account of which it has been added :
besides, the translation from the Arabic has this case explicitly de-
monstrated. And Proclus acknowledges, that the second part of
prop. 5, was added upon account of prop. 7, but gives a ridiculous
reason for it, ** that it might afford an answer to objections made
against the 7th," as if the case of the 7th, which is left out, were, as
he expressly makes it, an objection against the proposition itself. Who-
ever is curious, may read what Proclus says of this in his commen-
tary on the fifth and 7th propositions : for it is not worth whUe to re-
late his trifles* at full length.

It was thought proper to change the enunciation of this 7th prop,
so as to preserve the very same meaning ; the literal translation from
the Greek being extremely harsh, and difficult to be understood by
beginners.

PROP. XL B. I.

A corollary is added to this proposition, which is necessary to
prop. 1, B. 11, and otherwise.

PROP. XX. and XXI. B. I.

Proclus, in his commentary, relates, that the Epicureans derided
this proposition, as being manifest even to asses, and needing no de-
monstration ; and his answer is, that though the truth of it be mani-
fest to our senses, yet it is science which must give the reason why
two sides of a triangle are greater than the third: but the right an-
swer to this objection against this and the 21st and some other plain
propositions, is, that the number of axioms ought not to be increased
without necessity, as it must be if these propositions be not demon-
strated. Mons. Clairault, in the pre&oe to his Elements of Geometry,
published in French at Paris anno 1741, says, That Euclid has been
at the pains to prove, that the two sides of a triangle which is in-
cluded within another are together less than the. two sides of the tri-
angle which includes it ; but he has forgot to add this condition, viz.
that the triangles must be upon the same base : because, unless this
be added, the sides of the included triangle may be greater than the
sides of the triangle which includes it, in any ratio which is less than
that of two to one, as Pappus Alexandrinus has demonstrated, in
prop. 3. B. 3, of his mathematical collections.

PROP. xxn. B. L

Some authors blame Euclid, because he does hot demonstrate
that the two circles made use of in the cbnstruction of this pro-
blem must cut one another : but this is very [dain from the deter-
mination he has given, viz. that any tw« of the straight lines DF,

BOOK I.

NOTflS.

1141

FG, GH, must be greater than the
third : for who is so dull, though only
beginning to learn the Elements, as
not to perceive that the circle described
from the centre F, at the distance FD,
must meet FH betwixt F and H, be*
cause FD is less than FH; and that, DM FG H

for the like reason, the circle described from the centre G, at the
distance GH, or GM, must meet DG betwixt D and G ; and that
these circles must meet one another, because FD and GH are to-
gether greater than FG1 and this determination is easier to be un-
derstood than that which Mr. Thomas
Simson derives from it, and puts in-
stead of Euclid's In the 49th page of
his Elements of Geometry, that he
may supply the omission he blames
Euclid for; which determination is

that any of the three straight lines DM F G H

must be less than the sum, but greater than the difTerence of the
other two : from this he shows the circles must me^t one another^
in one case ; and says that it may be proved after the same manner-
in any other case ; but the straight line GM, which he bids take froiii>
GF, may be greater than it, as in the figure here annexed : in which
case hii^ depionstration must be changed into another.

PROP. XXIV. B. I.

To this is added, «• of the two sides DE, DF, let DE be that which
is not greater than the other;" that is, take that side of the two DE,*
DF, which is not greater than the other, in order to make with it
the angle EDG equal to BAG, because, without D
this restriction, there might be three different
cases of the proposition, as Campanusand others
make.

Mr. Thomas Simson, in p. 262 of the second
edition of his Elements of Geometry, printed
anno 1760, observes in his notes, that it ought
to have been shown that the point F fells below
the line EG. This probably Euclid omitted, as
it is very easy to perceive, that DG being equal g"
to DF, the point G is in the circumference of a
circle described from the centre D, at the dis-'
tance DF, and must be in that part of it which is above the straight
line EF, because DG falls above DF, the angle EDG being greater
than the angle EDF.

PROP, XXIX. B. I.

The proposition whidi is usually called the 5th postulate, or llth
axi(Hn, by some the twelfth, on which this 29th depends, has given a
great deal t%do, both to ancient and modem ge<»iieters : it seenuf

94ft vo^fifl. BOok I.

not to be properly placed among the axioms, as indeed it is not self-
evident ; but it may be demonstrated thus :

DEFINITiON I.

The distance of a point from a straight line,^ is the perpendicular
drawn to it from the point

DEF. 2.

One straight line is said to go nearer to, or further from another
straight line, when the distance of the pohits of the first from the
other straight Ihie becomes less or greater than they were ; and two
straight lines are said to keep the same distance from one another,
when the distance of the points of one of them from the other is al-
ways the same.

AXIOM.

A straight line cannot first come nearer to another straight line,
and then go further from it, before it cuts A -^--^-^^^ j_ — C

it ; and, in like manner, a straight line jy H . ^

cannot go further from another straight p — --..^^ _^^ • „

fine, and then come nearer to it ; nor can ^G

a straight line keep the same distance from another strai^t line,
and then come nearer to it, or go further from it ; for a straight line
keeps always the same direction.

For example the straight line ABC cannot fijrst come nearer to
the straight line DE, as from the point A B

to the point B, and then, from the point A ^^••— — > C

B to the point C, go further from the D E

same DE: and, in like manner the F ""q — - H

straight line FGH cannot go further from

DE, as from F to G, and then, from G to H, come nearer to the same

DE : and so in the last case, as in figure 2. (See the figure above.)

PROP. I.

If two equal straight lines, AC, BD, be each at right angles to the
same straight Ihie AB; if the points C, D be joined by the straight
1^ CD, the straight lin,e EF drawn from any point E in AB unto
CD, at right angles to AB, shall be equal to AC, or BD.

If EF be not equal to AC, one of them must be greater than the
other ; let AC be the greater ; then, because FE is less than CA, the
straight line CFD is nearer to the straight line AB at the point F
than at the point C, that is, CF comes F

nearer to AB firom the point C to F: but
because DB is greater than FE, the
straight line CFD is further from AB at
the point D than at F, that is FD goes
ftirther from AB from F to D : therefore
the straight line CFD first comes nearer

BOOK i. norm. iNS

to the straight line AB, and then goes farther from it, before ii eufs
It ; which is impossible. If FE be said to be greater than CA, or DB,
the straight line CFD first goes further from the straight line AB,
and then comes, nearer to it ; which is also impossible. Therefore
FE is not unequal to AC, that is, it is equal to it.

PROP. n.

♦

If two equal straight lines AC, BD be each at right angles to the
same straight line AB ; the straight line CD, which joins their ex-
tremities, makes right angles with AC and BD.

Join AD, BC ; and because, in the triangles CAB, DBA, CA, AB
are equal to DB, BA, and the angle CAB equal to the angle DBA;
the base BC is equal (4. 1.) to the base AD : and in the triangles
ACD, BDC, AC, CD, are equal to BD, DC, and the base AD is equal
to the base BC : therefore the angle ACD is 6 F D

equal (8. 1.) to the angle BDC: from any
point E in AB draw EF unto CD, at right
angles to AB : therefore by prop. 1. EF is
equal to AC, or BD ; wherefore, as has been
just now shown, the angle ACF is equal to A
the angle EFC : in the same manner, the angle BDF is equal to the
angle EFD ; but the angles ACD, BDC are equal ; therefore the an-
gles EFC and EFD are equal, and right angles (10. def. !.); where-
fore also the angles ACD, BDC are r^ht angles.

CoR. Hence, if two straight lines AB, CD be at right angles to the
same straight line AC, and if betwixt them a straight l&e BD be
drawn at right angles to either of them, as to AB; then BD I3 equal
to AC, and BDC is a right angle.

If AC be not equal to BD, take BG equal to AC, and join OG;
therefore, by this proposition, the angle, ACQ is a right angle ; but
ACD is also a right angle; wha*efore the angles ACD, ACG, are
equal to one another, which is impossible. Therefore BD is eqml to
AC; and by this proposition BDC is a right angle.

PROP. IlL

U two straight lines which contain an angle be produced, there
may be found in either side of them a point from which the perpeo-i
dicular drawn to the other shall be greater than any given straight
line.

Let AB, AC be two straight lines which make an angle with one
another, and let AD be the given straight line; a point may be found
either in AB ot AC, as in AC, from which the perpendicular drawn
to the other AB shall be greater than AD.

In AC take any point E, and draw EF perpendicular to AB ; pro-
duce AE to G, so that EG be equal to AE, and produce FE to H, and
make EH equal to FE, and join HG. Because in the triangles ABF,
GEH, AE, EF are equal to GE, EH, each to each, and contain equal
(15. 1.) angles, the angle GHE is therefore equal (4. 1.) to the angle
AFE, which is a right angle : draw GK perpendicular to AB ; and

944

iroTM.

BOOKi.

twcan^e the stmiglit Knesi FK, HG are at right aisled to FH, and

KG at right aisles to

F£;KGi0equaltoHF,

by cor. pr. 2* that is^ to

the double of EF. In the

same manner, if AG be

produced to L, so that D 4«

GL be equal to AG, and p J.

LM be drawn perpen- L

dieular to AB, then LM is double of GK, and so on. In AD take AN

equal to FE, and AO equal to KG, that is, to the double of FE, or AN ;

also take AP equal to LM, that is to the double of KG, or AO ; and,

kt this be done till the straight line taken be greater than AD ; let

this straight line so taken -be AP, and because AP is equal to LM,

therefore LM is greater than AD. Which was to be done.

PROP. IV.

If two straight lines AB, CD make equal angles BAB, ECD w^h
another straight line EAO towardsl the same parts of it ; AB and CD
lire at right angles to some straight line.

Bisect AC in F, and draw FG perpendicular to AB : take GH in
the straight line CD equal to AG, and on the contrary side of AC to
that on which AG is, and jcm FH : therdbre in the triangles AFG,
CFH, the sides FA, AG are equal to FC, CH, each to each, and the
angle FAG, that (15. 1 .) is EAB is equal
to the angle l^CH ; wherefore (4. 1.) the
angle AGF is equal to CHF, and AFG to
the ai^le CFH: to these last add the com-
a»oti angles AFH ; therefore the two an<*
^es AFG, AFH are equal to the two an-
gles CFH, HFA, which two last are equal
together to two right angles (13*1.), there-
fore also AFG, AFH are equal to two right

angles, and consequently (14. 1.) GF and q jj

FH are in one straight line. And because

AGF is a right angle, CHF which is equal to it is also a right angle ;

therefore the straight lines AB, CD are at right angles to GH.

PROP. V.

If two straight lines AB, CD, be cut by a third ACE so as to make
the interior angles BAC, ACD, on the same side of it, together less
than two right angles ; AB and CD being produced shall meet one
another towards the parts on which are the. two angles which are
less than two right angles.

. At the point C in the straight line CE make (23. 1.) the angle
ECF equal to the angle EAB, and draw to AB the straight line
CG at right angles to CF: then, because the angles ECF, EAB

BQOE U»

N9TBS.

345

are equal to one another, and
that the angles ECF, FCA are
together equal (13. 1.) to two
right angles, the angles £AB,
FCA are equal to two right
angles. But by the hypothe-
sis, the angles EAB, ACD are
together less than two right
angles; therefore the angle \ q (J
FCA is greater than ACD and
CD falls between CF and AB : and because CF and CD make an
angle with one another, by prop. 3, a point may be found in either
side of them CD, from which the perpendicular drawn to CF shall be
greater than the straight line CO. Let this point be H, and draw
HK perpendicular to CF, meeting AB in L : and because AB, CF
contain equal angles with AC on the same side of it, by prop. 4, AB
and CF are at right angles to the straight line MNO, which bisects
AC in N and is perpendicular to CF ; therefore, by cor. prop. 2, CG
and KL which are at right angles to CF are equal to one another;
and HK is ^eater than CG, aiid therefore is greater than KL, and
consequently the point. His jn KL produced. .Wherefore the straight
line GDH diHiWn betwixt the points C, H^ which are on contrary
sides of AL, must necessarily cut the straight Kne AL.

PROP. XXXV. R L

The demonstration of this proposition is changed, because, if the
method which is used in it was followed, there would be three cases
to be separately demonstrated, as is done in the translation from the
Arabic ; for, in the Elements, no case of a proposition that requires
a different demonstration! ought to be omitted. On this account we
have chosen the method which Mons. Clairault has given, the first
of any, as far as I know, in his Elements, page 21, and which afterr
wards Mr. Simson gives in his page 32. But whereas Mr. Simson
makes use of prop. 2(5, b. 1, from which the equality of the two tri-
angles does not immediately follow, because, to prove that, the 4th
of b. 1, must likewise be made use of, as may be seen in the very
same case in the 34th prop. b. I, it was thought better to make use
only of the 4th of b. 1.

PROP. XLV. B. L

The straight line KM is proved to be parallel to FL from the 33d
prop. : whereas KH is parallel to FG by construction, and KHM»
FGL, have been demonstrated to be straight lines. A coroUaiy is
added from Commandine, as being often used.

* PROP. xm. B. n.

lu this proposition only acute angled triangles are mentionedt
whereas it holds true of every triangle ; and the demonstrati(»is of
the cases on)iUed are added : Commandine and Clavlus have like*
wise given their demonstrations of these cai^es.

246 NOTES. B6OKIIT.

PROP. XIV. B. 11.

In the demonstration of this, some Greek editor has ignorantly in-
serted the words " but if not, one of the two BE, ED is the greater ;
let BE be the greater, and produce it to F," as if it was of any con-
sequence whether the greater or lesser be produced : therefore, ia-
stead of these words, there ought to be read only, " but if not, pro-
duce BE to F."

PROP. I. B. in.

Several authors, especially among the modern mathematicians and
logicians, inveigh too severely against indirect or apagogic demon-
strations, and sometimes ignorantly enough, not being aware that
there are some things that cannot be demonstrated any other way :
of this the present proposition is a very clear instance, as no direct
demonstration can be given of it : because, besides the definition of
a circle, there is no principle or property relating to a circle antece-
dent to this problem, from which either a direct or indirect demon-
stration can be deduced : wherefore it is necessary that the point
found by the construction of the problem be proved to be the centre
of the circle, by the help of this definition, and 'some of the preceding
propositions : and because, in the demonstration, this proposition
must be brought in, viz. straight lines from the centre of a circle to
the circumference are equal, and that the point found by the con-
struction cannot be assumed as the centre, for this is the thing to be
demonstrated ; it is manifest some other point must be assumed as
'the centre; and if from this assumption an absurdity follows, as
Euclid demonstrates there must, then it is not true that the point as-
sumed is the centre ; and as any point whatever was assumed, it
follows that no point, except that found by the construction, can be
the centre, from which the necessity of an indirect demonstration in
in this case is evident

PROP. Xffl. B. m.

As it is much easier to imagine that two circles may touch one
another within, in more points than one, upon the same side, than
upon opposite sides ; the figure of that case ought not to have been
omitted ; but the construction in the Greek text would not have
suited with this figure so well, because the centres of the circles
must have been placed near to the circumferences ; on which ac-
count another construction and demonstration is given, which is the
same with the second part of that which Campanus has translated
from the Arabic, where, without any reason, the demonstration is
divided into two parts.

PROP. XV. B. ni.

The converse of the second part of this proposition is wanting,
though in the preceding, the converse is added, in a like case,
both in the enunciation and demonstration; and it is now added
in this. Besides, in the demonstration of the first part of this

BOOK UL NOTES. %417

15tb, the diameter AD (see Commandine's figure) is proved to be
greater than the straight line BC, by means of another straight
line MN; whereas it may be better done without it: on which
accounts we have given a different demonstration, like to that
which Euclid gives in the preceding I4th, and to that which
Theodosius gives in prop. 6, b. 1, of iiis Spherics in this very
affair.

PROP. XVI. B. m.

In this we have not followed the Greek nor the Latin transla-
tion literally, but have given what is plainly the meaning of this
proposition, without mentioning the angle of the semicircle, or
that which some call the comicular angle, which they conceive
to be made by the circumference and the straight line which is at
right angles to the diameter, at its extremity; which angles have
furnished matter of great debate between some of the modern
geometers, and given occasion of deducing strange consequences
from them, which are quite avoided by the manner in which we
have expressed the proposition. And in like manner, we have
given the true meaning of prop. 31, b. 3, without mentioning the
angles of the greater or lesser segments: these passages Vieta,
with good reason, suspects to be adulterated, in the 386th page
of his Oper. Math.

PROP. XX. k in.

The first words of the second part of this demonstration;
•^ x«xXa$^w Svi *aXiv," are wrong translated by Mr. Briggs and Dr.
Gregory "Rursus inclinetur;" for the translation ought to be
**Rursus inflectatur;'* as Commandine has it: a straight line is
said to be inflected either to a straight or curve line, when a
straight line is drawn to this line from a point, and from the
point in which it meets it, a straight line making an angle with
the former is drawn to another point, as is evident from the 90th>
prop, of Euclid's Data: for this the whole line betwixt the first
and last points, is inflected or broken at the point of inflection,
where the two straight lines meet. And in the like sense two
straight .lines are said to be inflected from two points to a third
point, when they make an angle at this point ; as may be seen in
the description given by Pappus Alexandrinus of Apollonius's
books de Locis planis, in the preface to his 7th book: we have
made the expression fuller from the 90th prop, of the Data.

PROP. XXI. B. m.

There are two cases of this proposition, the second of which,
viz. when the angles are in a segment not greater than a semicir-
cle, is wanting in the Greek : and of this a more simple demon-
stration is given than that. which is in Commandine, as being de-
rived only firom the first case, without the help of triangles.

PROP. XXIII. andX:XIV. B. III.

In proposition 24 it is demonstrated, that the segment AEB
must coincide with the segment CFD, (see Gommandine's figure)
and that it cannot fall otherwise, as OGrD, so as to cut the other

NOTES. BOOK Itl.

d^de in a third point G, from this, that, if it did, a cirde cotdd
cut another in more points than two : but this ought to have been
proved to be impossible in the 23d prop, as well as that one of
the segments eannot &11 within the other : this part then in left
out in the 24th, and put in its proper place, the 23d proposition.

PROP. XXV. B. m. .

This proposition is divided into three cases, of which two have
the same construction and demonstration ; therefore it is now di-
vided only into two cases.

PROP. xxxm. B. in.

This also in the Greeic is divided into three cases, of which
two, viz. one in which the given angle is acute, and the other in
which it is obtuse, have exactly the same construction and de-
monstration; on which account, the demonstration of the last
case is left out as quite superfluous, and the addition of some un-
skilful editor ; besides, the demonstration of the case when the
angle given is a right angle, is done a round about way, and is
therefore changed to a more simple one, as was done by Clavius. ,

PROP. XXXV. B. III.

As the 25th and 33d propositions are divided into more cases,
so this thirty-fifth is divided into fewer cases than are necessary.
Nor can it be supposed that Euclid omitted thein because they
are easy ; as he has given the case, which by far is the easiest of
them all, viz. that in which both the straight lines pass through
the centre; and in the following proposition he separately de-
monstrates the case in which the straight line passes through the
centre, and that in which it does not pass through the centre : so
that it seems Theon, or some other, has thought them too long
to insert : but cases that require different demonstrations, should
not be left out in the Elements, as was before taken notice of:
these cases are in the translation from the Arabic, and are now
put into the text.

PROP. XXXVII. B. in.'

At the end of this the words ** in the same mjanner it may be
demonstrated, if the centre be in AC," are left out as the addition
of some ignorant editor.

DEFINITIONS OF BOOK IV.

When a point is in a straight line, or any other line, this point
is by the Greek geometers said azsrsif^aif to be upon, or in that
line, and when a straight line or circle meets a circle any way,
the one is said airrsd^ai, to meet the other: but when a straight
line or circle meets a circle so as not to cut it, it is said 6(paitrsff^ai,
to touch the circle : and these two terms are never promiscuously
used by them: therefore in the fifth defimtion of book 4» the com-
pound «96Mrri)rai must be read, instead of the simple a^fiKai; and

BOOK V. MOTES. t4l9

in the 1st, 3d, 3d, and eth definitions in Commandine's transla-
^on, "tangit,'* must be read instead of «contingit;" and in the 2d
and 3d definitions of book 8, the same change must be made : bnC
in the Oreelc test of propositions 11th, 12th, 18th, 18th l»th,
book 8, the eomponnd verb is to be pat for the simple.

PROP. IV. B.IV.

In thi8» as also in the 8th and 13th j>ropositions of this book, it
is demonstrated indirectly, that the circle touches a straight line :
whereas in the 17th» 33d, and 37th proportions of book 3y the
same thing is directly demonstrated : and this way we have chosen
to use. in the propositions of this book, a$ it is Sorter.

PROP. V. B.IV.

The demonstration of this has been spoiled by some unsldlfu^
hand: lor he does not demonstrate, as is necessary, that the two
straight lines which bisect the sides of the triangle at right angles
must meet one another; and, without any reason, he dit^ldes the
proposition into three cases; whereas;, one and the s^me construc-
tion and demonstration serves for them all, as Campanus has oi>"
served; which useless repetitions are now left out: the Greek
text also in the corollary is man^stly vitiated, where mention is
fnade of a given angle, though there neither kf nor ean be any
fhkig in the preposition relating to a given angle.

PROP. XV. and XVL B. IV.

In the corollary of the first of these, the words equflateral and
^[|uiangular are wanting in the Greek: and in prop, 16, instead
of the clrcie ABCD, ought to be read the circumference ABCD:
where mention is made of its containing fifteen equal parts.

DEP. m. B. V.

Many of the modern mathematickns reject this definition : the
very learned Dr. Barrow has explained it at large at the end of
hrs third lecture of the year 166&; in whfch also he answers the
objections made against it as well as the subject would afUow: and
at the end gives his opinion upon the whole as follows:

" I shall only add, that the author had, perhaps^ no other design
in makbag this definition, than (that he might more fully explain
and embellish his subject) to give a general and summary idea of
ratio to beginners, by premising this metaphysical definition, to
the more accurate definitions of rafios that are the same to one
another, or one of which is greater^ or fess than Che other; I call
it a metaphyseal, for it is not properly a merthematk^ definition,
sdnee nothing in mathematics depends on it, or is deduced* nor,
as I judge, can he deduced from it ; and the definition of analogy*
wMch follows, viz. Analogy is the similitude of ratios* is of the
same kind, and can serve for no purpose in mathematics, but only
to give beginners some general, though gross and confused notions
of analogy : but the whole of the doctrine of ratios, and the whole
of mathematics, depend upon the accurate mathematical defini*

260 Ncyrca book v.

tions which follow this : to these we ought prindpaUy to aitend,
as the doctrine of ratios is more perfectly explained by them:
this third and others lilce it, may be entirely spared without any
loss to geometry: as we see in the 7th boc^ of the ESements,
where the proportion of numbers to one another is d^ned, and
treated of, yet without giving any definition of the ratio of num-
bers; though such a definition was as necessary and useful to be
given in ttmt book, as in thia : but indeed there is scarce any need
of it in either of them : though I think that a thing of so general
and abstracted a nature, and thereby the more difficult to be con-
ceived an4 explained, cannot be more commodiously defined than
as the author has done; upon which account I thought fit to ex-
plain it at large, and defend it against the captious objections of
those who attack it." To this citation from Dr. Barrow I have
nothing to add, except that 1 fully believe the 3d and 8th defini-
tions are not EucIid^s, but added by some unskilful editor.

DEP. XL R V.

It was necessary to add the word "continual" before "propor-
tionals" in this definition ; and thus it is cited in the 33d prop, of
book 11.

After this definition ought to have followed the definition of
compound ratio, as this was the proper place for it ; duplicate and
triplicate ratio being species of compound ratio. But Theon has
made it the 5th def. of book 6, where he gives an absurd and en-
tirely useless definition of compound ratio : for this reason we have
placed another definition of it betwixt the 11th and 12th of 'this
book, which, no doubt, Euclid gave; for he cites it expressly in
prop. 23, book 6, and which Clavius, Herigon, and Barrow have
likewise given, but they retain also Theon^s, which they ought to
have left out of the Elements.

DEF. XIII. B. V.

This, and the rest of the definitions following, contain the ex-
plication of some terms which are used in the 6th and fbllowmg
books; which, except a few, are easily enough understood from
the propositions of this book where they are first mentioned : they
seem to have been added by Theon, or some other. However it
be, they are explained something more distinctly for the sake of
learners,

PROP. IV. B. V.

In the construction preceding the demonstration of this, the
words a STvxs, any whatever, are twice wanting in the Greek, as
also in the Latin translations; and are now add^, as being whdly
necessary.

Ibid, in the demonstration; in the Greek, and in the Latin trans-
lation of Commandine, and in that of Mr. Henry Briggs, which was
published at London in 1620, together with the Greek text of the
first six books, which translation in this place is followed by Dr.
Gregory in his edition of Euclid, there is this sentence following,
viz. "and of A and C have been taken equimultiples K, L; and of

BOOK V. NOTES. 261

B and D, any equimultiples whatever (a sruxO M, N f which is not
true, the words " any whatever," ought to be left out: and it is strange
that neither Mr. Briggs, who did right to leave out these wordi| in
one place of prop. 13, of this boolc, nor Dr. Gregory, who changed
them into the word ** some," in three places, and left them out in a
fourth of that same prop. 13, did not also leave them out in this place
of prop. 4y and in the second of the two places where they occur in
prop 17, of this book, in neither of which they can stand consistent
with truth: and m none of all these places, even in those which they
corrected in their Latin translation, have they cancelled the words*
a STvxs in the Greek text, as they ought to have done.

The same words a srvxs are found in four places of prop. 11, of
this book, in the first and last of which they are necessary, but in the
second and third, though they are true, they are quite superfluous ;
as they likewise are in. the second of the two places in whicl\ they
are found in the 12th prop, and in the like places of prop. 22, 23 of
this book; but are wanting in the last place of prop. 23, as also in
prop. 25, book 11.

COR. IV. PROP. B. V.

This corollary has beai unskilfully aimexed to this proposition^
and has been made instead of the legitimate demonstration, which,
without doubt, Theon, or some other editor, has taken away, not from
this, but from its proper place in this book : the author of it designed
to demonstrate, that if four magnitudes E, G, F, H be proportionals,
they are also proportionals inversely, that is, G is to E, as H to F ;
which is true ; but the demonstration of it does not in the least de-
pend upon this 4th prop, or its demonstration : for, when he says,
" because it is demonstrated that if K be greater than M, L is great-
er than N," &xi, — This indeed is shown in the demonstration of the
4th prop, but not from this, that E, G, F, H are proportionals : for this
last is the conclusion of the proposition. Wherefore these words,
" because it is demonstrated," &c. are wholly foreign to his design ;
and he should have proved, that if K be greater than M, L is greater
than N, from this, that E, G, F, H are proportionals, and from the 5th
def. of this book, which he has not ; but is done in proposition B,
which we have given^in its proper place instead of this corollary; and
another corollary is placed after the 4th prop, which is often of use ;
and is necessary to the demonstration of prop. 18. of this book.

PROP. V. B. V.

In the construction which precedes the demonstration of this pro-
position, it is required that EB may be the same multiple of CG, that
AE is of CF; that is, that EB be divided into as many equal parts,
as there are parts in AE equal to CF : from which it is evident, that
this construction is not Euclid's ; for he does not show the way of
dividing straight lines, and for less other magnitudes, into any num-
fo^ (^ equal parts, until the 9th proposition of book 6 ; and he never
requires any thing to be done in the construction, of which he had

F»

9M MOTES. BOQIC V.

not before given the method of doing. For this rea-
gon, we have changed the construction to one, which, A
without doubt, is Euclid's, in which nothiagds required | G
bur to add a magnitude to itself a certain number of ^"
times ; and this is to be found in the translation from q.

the Arabic, though the enunciation of the proposition
and the demonstration are there very much spoiled.
Jacobus Peletarius, who was the first, as &r as I know, ^
who took notice of this error, gives also the right con* ^

•struction in his edition of £uclid| after he had given
the other which he blames. He says, he would not leave it out, be-
cause it was fine, and might sharpen one's genius to invent others
like it ; whereas, there is not the least difference between the two
demonstrations, except a single word in the construction, which
very probably has been owing to an unskilful librarian. Clavius
likewise gives both the ways ; but neither he nor Pdetarius, takes
notice of the reason why the one is preferable to the other.

PROP. VI. B. V.

There are two cases of this proposition, of which only the first and
simplest is demonstrated in the Greek : and it is probable Theon
thought it was sufficient to give this one, since he was to make use
of neither of them in his mutilated edition of the 8th book ; and he
might as well have 1^ out the other, as also the dth proposition, for
the same reason. The demonstration of the other case is now added,
because both of them, as also the dth proposition, are necessary to the
demonstration of the 18th proposition of this book. The translation
from the Arabic gives both cases briefly.

PROP. A. B. V.

This proposition is frequently used by geometers, and it is neces*
sary in the 25th prop, of this book, 31st of the 6th, and 34th of the
1 1th, and 15th of the 12th book. It seems to have been taken out
of the Elements by Theon, because it appeared evident enough to
him, and others, who substitute the confused and indistinct id^ the
vulgar have of proportionals, in place of that accurate idea, which is
to be got from the 5th definition of this book. Nor can there be any
doubt that Eudoxius or Euclid gave it a place in the Elements, when
we see the 7th £^d 9th of the same book demonstrated, though they
are quite as easy and evident as this. Alphonsus Borellus takes oc-
casion from this proposition to censure the 5th definition of this book
very severely, but most unjustly. In p. l.Sd6, of his EuoUd restored,
{urlnted at Pisa in 1658, he says, *' Nor can even this least deigree of
knowledge be obtained firom the aforesaid property,^' viz, that whx3h
is contained in 5th def. 5. "That if four magnitudes be prc^ixv-
tionals, the third must necessarily be greater than the.foarth^ when
the first is greater than the second; as Clavius aoknowledlges in the
16th prop, of the 5th book of the Elements." But though Clavius
makes no such acknowledgment expressly, he has given BoreHas
a handle to say this of him : because when Clavius, in the above

BOOST.

NOTB8.

2§d

cited plaoe, censures CkHnmandinet and that very justly, lor demon-
strating this proposition by help of the 16th of the fifth ; yet he him-
self gives no demonstration of it, but thinks it |dain from the naUire
of proportionals, as he writes in the end of the 14th and 16th prc^.
book 5, of his edition, and is followed by Herigon in Schol. 1, prop.
14tb, book 5, as if there was any nature of proportionals antecedent
to that which is to be derived and understood from the definition of
them. And« indeed, though it is very easy to give a right demon-
stration of it, nobody, as ^r as I know, has given one, except the
learned Dr. Barrow, who, in answer to Borellus's objection, demon-
strates it indirectly, but very briefly and clearly, from the 5th defi-
nition in the 322d page of his Lect. Mathem. from which definitioft
it may also be easily demonstrated directly. On which account we
have placed it next to the propositions concerning equimultiples.

PROP. B. B. V.

This also is easily deduced from the 5th def. b. 5, and therefore is
placed next to the other ; for it was very ignorantly made a corollary
from the fourth prop, of this book. See the note on that cctt'ollary.

PROP. C. B. V,

This is frequently made use of by geometers, and is necessary to •
the 5th and 6th propositions of the 10th book : Clavius, in his notes
subjoined to the 8th def. of book 5, demonstrates it only in numbers*
by help of some of the propositions of the 7th book : in order to de^
monstrote the property contained in the 5th definition of the 5th
book, when aj^lied to numbers, from the property of proportionals
contained in the 20th def <^ the 7th book. And most of the com-
mentators judge it difficult to prove that four magnitudes which are
proportionaJs according to the 20th def of 7th book, are also propor-
ttonals according to the 5th def <^ 5th book. But this ia easOy made
out, as foUows :

First, If A, B, C, D, be four magm- F

tudes, such that A ia^ibe same multiple, B H

or the saone pan of B, whwhOisofD;
A, B, C, D are proportionals. This is
demonstrated in proposition C.

Secondly, if AB contain -the same
parts of CD, that EF does of GH ; in
this case likewise AB is to CD, as £F
toGH. A G & Q

Let CK be a part of CD, and GL the same part of GH,and let AB
be the same multiple of CK, that £F is
of OL : therefore, by prop. C of 5th B
book, AB is to CK, as EF to GL : and D

CD, GH are equimultiples of CK, GL the H

second and (Gburth : wherefi^re by eon
pn^. 4, book 5, AB Is to CD» as SF to
GH.

And if fear magnitodes be propor-
tionals according to the 5th def. of A Cl &

K—

254

KOTEft.

BO«K V.

book 6, they are also proportionals according to the 2(N;h def. of
book 7.

First, if A be to B, as C to D ; then if A be any multiple or part
of B, C is the same multiple or part of D, by prop. D, of book 5.

Next, if AB be to CD, as EF to GH ; then if AB contains any
parts of CD, EF contains the same parts of GH : for let CK be a part
of CD, and GL the same part of GH, and let AB be a multiple of
CE ; EF is the same multiple of GL ; take M the same jnultiple of
GL that AB is of CK ; therefore by prop. C, of book 5, AB is to CK,
as M to GL ; and CD, GH are equimultiples of CK, GL : wherefore
by cor. prop. 4, b. 5, AB is to CD, as M to GH. And, by the hypo-
thesis, AB is to CD as EF to GH : therefore M is equal to EF, by
prop. 9, book 5, and consequently EF is the same multiple of GL that
ABisofCK.

PROP. D. B. V.

This is not unfrequently used in the demonstration of other propo-
sitions, and is necessary in that of prop. 9, b. 6. It seems Tbeon
has left it out for the reasons mentioned in the notes of prop. A.

PROP. vra. B. V.

In the demonstration of this, as it is now in the Greek, there are
two caises, (see the demonstration in Hervagius, or Dr. Gregory's
edition,) of which the first is that in which AE is less than EB ; and
in this it necessarily follows, that H0 the multiple of EB is greater
than ZH, the same multiple of AE, which last multiple, by the con*
struction is greater than A ; whence also H0 must be greater than
A, But in the second case, viz. that in which EB is less than AE,
though ZH be greater than A, yet H0 may be less than the same
A ; so that there cannot be taken a multiple of A which is the first
that is greater than K, or H0 because a itself is greater than it ;
upon this account the author of this demonstration found it neces-
sary to change one part of the construction that was made use of
in the first case : but he has, without any necessity, changed also
another part of it, viz. when he orders to take N that multiple of
A which is the first *that is greater "~

than ZH ; for he might have taken
that multiple of A. which is the first
that is greater than H0, or K, as
was done in the first case^ he
likewise brings in this K into the
demonstration of both cases, with-
out any reason ; for it serves to no
purpose but to lengthen the demon-
stration. There is also a third
case, which is not mentioned in

this demonstration, viz. that in which AE in the first, or EB in the
second of the two other cases, is greater than D; and in this fmy
equimultiples, as the doubles, of AE, KB are to be taken, as is done
in this edition, where all the cases are at once demonstrated : and

H-

1
A

H—

B A

BOOK V. NOTES. 2&5

from this it is plain tiiat Tlieon, or some other unskiJful editor has
vitiated this proposition. p

PROP. IX. B. V.

Of this there is given a more explicit demonstration than that which
is now in the Elements.

PROP. X. B. V.

It was necessary to give another demonstration of this proposition,
because that which is in the Greek and Latin, or other editions, is
not legitimate : for the words greater, the same, or equal, leaser, have
a quite different meaning when applied to magnitudes and ratios, as
is plain from the 5th and 7th definitions of boolc 5. By the help of
these let us examine the demonstration of the 10th prop, which pro-
ceeds thus : " Let A have to C a greater ratio than B to C: I say that
A is greater than B. For if it is not greater, it is either equal, or less.
But A cannot be equal to B, because then each of them would have
the same ratio to C : but they have not. Therefore A is not equal to
R" The force of which reasoning is this ; if A had to C the same
ratio that B has to C ; then if any equimultiples whatever of A and B
be taken, and any multiple whatever of C ; if the multiple of A be
greater than the multiple of C, then, by the 5th def. of book 5, the
multiple of B is also greater than that of C ; but, from the hypothesis
that A has a greater ratio to C, than B has to C, there must, by the
7th def. of book 5, be certain equimultiples of A and B, and some
multiple of C, sudi that the multiple of A is greater than the multi-
pie of C, but the multiple of B is not greater than the same multiple
of C: and this proposition directly contradicts the preceding: where-
fore A isnot equal to B. The demonstration of the 10th prop, goes
thus : ** but neither is A less than B ; because then A would have a
less ratio to C than B has to it : but it has not a less ratio, therefore
A is not less than B," &c. Here it i^ said, that *^ A would have a
less ratio to than B has to C* or which is the same thing, that B
would have a greater ratio to C than A to C ; that is, by the 7th def.
book 5, there must be some equimultiples of B and A, and some mul-
tiple of C, such that the multiple of B is greater than the multiple of
Cy but the multiple of A is not greater than it : and it ought to have
been proved, that this can never happen if the ratio of A to C be
greater than the ratio of B to C ; that is, it should have been proved,
that in this case, the multiple of A is always greater than the multi-
ple of C, whenever the multiple of B is greater than the multiple of
; for when this is demonstrated, it will be evident that B cannot
have a greater ratio to C, than A has to C, or, which is the same
thing, that A cannot have a less ratio to C than B has to C : but this
is not at all proved in the 10th proposition ; but if the 10th were once
demonstrated, it would immediately follow from it, but cannot with*
out it be easily demonstrated, as he that tries to do it will find.
Wherefore the 10th proposition is not sufficiently demonstrated.
And it seems that he who has given the demonstration of the 10th
proposition as we now have it» instead of that which Eudoxus or

256

NOTES.

BOOK V.

•

1
1

A C B C

D F E F

Buclid had given, has been deceived in applying what is manifest,
when understood of magnitudes, unto ratios, viz. that a magnitude
cannot be both greater and less than another. That those things
which are equal to the same are equal to one another, is a most
evident axiom when understood of magnitudes ; yet Euclid does not
make use of it to infer that those ratios which are the same to the
same ratio, are the same to one another ; but explicitly demonstrates
this in prop. 11, of book &. The demonstration we have given of
the 10th prop, is no doubt the same with that of Eudoxus or Euclid,
as it is immediately and directly derived from the definition of a
greater ratio, viz. the 7th of the 5th,

The above mentioned proposition, viz. If A have ta C « greater
ratio than B to C : and if of A and B there be
taken certain equimultiples, and some multiple
of C ; then if the muMpie of B be greater than
the multiple of C, the multiple of A is also great-
er than the same, is thus demonstrated.

Let D, E be the equimultiples of A, B,,and F
a multiple of C, such, that E the multiple of B
is greater than F ; D the multiple of A is also
greater than F.

Because A has a greater ratio to C, than B
to C, A is greater than B, by the 10th prop,
book 5 ; therefore D the multi{de of A is great*
er than E the same multiple of B : and E is
greater than F; much more therefore D is
greater than F.

PROP. xm. B. V.

In Cbmmandine's, Briggs's, and Gregory's translations, at the be-
ginning of this demonstration, it is said, *^ And the multiple of C is
greater than the multiple of D ; but the multiple of E is not greater
than the multiple of F ;" which words are a literal translation from
the Greek ; but the sense evidently requires that it be read, " so that
the multiple of C be greater than the multiple of D; but the multiple
of E be not greater than the multiple of F." And thus this place was
restored to the true reading in the first editions of Commandlne's
Euclid, printed in 8vo. at Oxford ; but in the later editions, at least
in that of 1747, the error of the Greek text was kept in.

There is a corollary added to prop, 13, as it is necessary to the
20th and 2 1st prop, of this book, and is as useful as the proposition.

PROP. XIV. B. V.

The two cases of this, which are not in the Greek, are added ; the
demonstration of them not being exactly the same with that of the
first case.

PROP. XVII. B. V.

The order of the words in a clause of this is changed to one more
natural; as was idso done in prop. 1.

;

i 1

'■ 1

( \

BO<Hc V. iroTia. W7

PROP. xvm. B. V.

The demoQAtvatiQn of this is none of Euclid's, nor is it legitimate ;
for it depends upon this hypothesis, that to any three magmtudes*
two oi which, at least, are of the same kind, there may be a fourth
proportional : which, if not proved, the demonstration now in the text
is of no force : but this is. assumed without any proof; nor can it as
far as I am able to discern, be demonstrated by the propositions pre-
ceding this : so far is it from deserving to be reckoned an axiom, as
Clavius, after other commentators, would have it, at the end of the
definitions of the 5th book. Euclid does not demonstrate it, nor does
he show how to find the fourth proportional, before the 12th prop, of
tbe sixth book : and he n9ver assumes any thing in the demonstra-
tion of a pr(^x>sition, which he had not before demonstrated : at least,
he assumes nothing the existence of which is not evidently possible;
for a certain conclusion can. never be deduced by the means of an
uncertain proposition : upon this account, we have given a legitimate
demonstration of this proposition instead of that in the Greek and
other editions, which very probably Theon, at least some other, has
put in the place of Euclid's, because he thought it too prolix : and as
tiie 17th prop, oi which this 18th is the converse, is demonstrated by
help a( the first and second fNPopositioas of this book ; so, in the de-
monstration now given of the 18th, the 5th prop, and both eases of
the 0th are necessaury, and these two propositions are the converse
of the 1st and 2d. Now the 5th kad 6th do not eotet into the de-
monstration of any proponitioa in this book as we now have it : nor
can they be of use in any proposition of the Elements, except in thia
18th, and this is a nmniiest prooC that Euclid made use of them in
his demonstration of it, and that the demonstration now given, which
is exactly the converse of that of the 17th, as it ought to be, differs
nothing from that of Eudoxus or Euclid : for the 5th and 6th have
undoubtedly been put into the 5th book for the sake oi some propose
ticms in it, as all the other propositions about equimultiples have been.

Hienmymus Saccherius, in his book named Euclides ab omni nsBvo
vindicatus, printed at Milan, anno 1733, in 4to» acknowledges this
blemish in the demonstration of the 18th, and that he may remove it,
and render the demonstration we now have of it legitimate, he en-
deavours to demonstrate the foUowhig proposkion, which is in page
115 of his book, viz.

'^ Let A,.Bk C, D be four magnitudes, of which the two first are of
the one kind, and also the two others dther of the same kind with the
two first, or of some other, the same kind with one another, I say
the ratio of the third C to the ^urth D, is either equal to, or greater^
or less than the ratio of the first A to the second B."

And alter two propositions premised as lemmas, he proceeds
thus:

^ Either among all the possible equimultiples of the first A, and of
the third C, and at the same time, among all the possible equimulti-
ples of the second B, and of the fourth D, there can be found some
one multiple EF of the first A, and one IK of the second B, that are

258 NOTES. BOOK V.

equal to one another ; and also, in the same case, some one multiple
GH of the third C equal to LM the multiple of the fourth D, or such
equiality is no where to be found. If the first case happen [i. e. if
such equality is to

be found] it is mani- A E— — F

fest from what is be-
fore demonstrated, B— — I K
that A is to B, as C

to D ; but if such si- C G-^ ^H

multaneous equality

be not to be found D L M

upon both sides, it

will be found either upon one side, as upoB the side of A [and B;] or
it will be found upon neither side ; if the first happen ; therefore (fi*om
Euclid's definition of greater and lesser ratio foregoing) A has to B
a greater or less ratio than C to D ; according as GH the multiple of
the third C is less, or greater than LM the multiple of the fourth D :
but if the second case happen : therefore upon the one side, as upon
the side of A the first and B the second, it may happen that the mul-
tiple EF, [viz. of the first] may be less than IK the multiple of the
second, while, on the contrary, upon the other side, [viz. of C and D]
the multiple GH [of the third C] is greater than the other multiple
LM [of the fourth D :] and then (from the same definition of Euclid)
the ratio of the first A to the second B, is less than the ratio of the
third C to the fourth D ; or on the contrary.

" Therefore the axiom [i. e. the proposition before set down] re-
mains demonstrated," &c.

Not in the least ; but it remains still undemonstrated ; for what he
says may happen, may, in innumerable cases never happen; and
therefore his demonstration does not hold : for example, if A be the
side, and B the diameter of a square; and C the side, and D the di-
ameter of another square; there can in no case be any multiple of A
equal to any of B ; nor any one of C equal to one of D, as is well
known ; and yet it cam never happen, that when any multiple of A
is greater than a multiple of B, the multiple of C can be less than the
multiple of D, nor when the multiple of A is less than that of B, the
multiple of C can be greater than that of D, viz. taking equimultiples
of A and C, and equimultiples of B and D : for A, B, C, D are pro-
portionals ; and so if the multiple of A be greater, &c. than that of Br
so must that of C be greater, &c. than that of D; by 5th def. b. 5.

The same objection holds good against the demonstrations which
some give of the 1st prop, of the 6th book, which we have made
against this of the 18th prop, because it depends upon the same in-
sufficient foundation with the other.

PROP. XIX. B. V.

A corollary is added to this, which is as frequently used as the
proposition itself. The corollary which is subjoined to it in the
Greek, plainly shows that the 5th book has been vitiated by edi-

soar T. MOTES. 3S0

tors who were not geometers : for the conversion of ratios does not
depend upon this 19th, and the demonstration which several of the
commentators on Euclid give of conversion is not legitimate, as Cla-
vlus has rightJy observed, who has given a good demonstration of
it, which we have put in proposition E ; but he makes it a corollary
from the 19th, and begins it with the words, "Hence it easily fol-
lows," though it does not at all follow from it.

PROP. XX. XXI. XXII. xxin. XXIV. B. V.

The demonstrations of the 20th and 21st propositions, are shorter
than those Euclid gives of easier propositions, either in the preceding
or following books : wherefore it was proper to make them more
explicit, and the 22d and 2dd propositions are, as they ought to be,
extended to any number of magnitudes : and, in like manner may
the 24th be, as is taken notice of in a corollary ; and another
corollary is added, as useful as the proposition, and the words " any
whatever" are supplied near the end of prop. 23, which are wanting
in the Greek text, and the translations from it.

In a paper writ by Philippus Naudaeus, and published after his
death, in the History of the Royal Academy of Sciences of Berlin,
anno 1745, page 50, the 23d prop, of the 5th book is censured as
being obscurely enunciated, and, because of this, prolixly demon-
strated : the enunciation there given is not Euclid's, but Tacquet^s,
as he acknowledges, which, though not so well expressed, is, upon
the matter, the same with that which is now in the Elements. Nor
is there any thing obscure in it, though the author of the paper has
set down the proportionals in a disadvantageous order, by which it
appears to be obscure : but, no doubt, Euclid enunciated this 23d, a^
well as the 22d, so as to extend it to any number of magnitudes,
which taken two and two are proportionals, and not of six only; and
to this general case, the enunciation which Naudaeus gives, cannot be
well applied.

The demonstration which is given of this 23d, in that paper, is quite
wrong ; because, if the proportional magnitudes be plane or solid
figures, there can no rectangle (which he improperly calls a product)
be conceived to be made by any two of them, and if it should be
said that in this case straight lines are to be taken which are pro-
portional to the figures, the demonstration would this way become
much longer than Euclid's : but, even though his demonstration had
been right, who does not see that it could not be made use of in the
5th book?

PROP. P, G, H, K. B. V,

These propositions are annexed to the 5th book, because they
are frequently made use of by both ancient and modern geometers :
and in many cases compound ratios cannot be brought into de-
monstration, without making use of them.

Whoever desires to see' the doctrine of ratios to be delivered in this
5th book solidly defended, and the arguments brought against it by
And. Tacquet, Alph. Borellus, and others, fully refuted, may read

900 NOTBS. B00& TI.

Dr. Banrow's Mathematical Lectures, viz. the 7th andSthof tiie year
1666.

The 5th book being thus corrected, I most readily agree to what
tl)e learned Dr. Barrow says,* '* That there is nothiug in the whole
body of the Elements of a more subtle invention, nothing more solid-
ly established, and more accurately handled, than the doctrine of
proportionals." And there is some ground to hope, that gecnneters
will think that this could not have been said with as good reason,
since Theon's time till the present.

DEF. n. and V. of B. VI.

The 2d definition does not seem to be Euclid's, but some unskilful
editor's : for there is no mention made by Euclid, nor, as &r as I
know, by any other geometer^ of reciprocal figures: it is obscurely
escpressed, which made it prop^ to raider it mare distinct : it would
be better to put the following defimtion in place of it, viz.

DEF. fl.

Two magnitudes are said to be reciprocally proportional to two
others, when one of the first is to one of the other magnitudes, as the
remahiing one of the last two is to the remaining one of the first

But the filth definition, which, suice Theon's time, has been kept
in the Elements, to the great detriment of learners, is now Justly
thrown out of them, for the reason given in the notes on the 23d
prop, of this book.

PROP. L and H. B. VI.

To the first o( these a cordlary is added, which is often used : and
^e enunciation of the second is made more general.

PROP. m. B. .VI.

A second case of this, as useful as the first, is* given in prop. A :
viz. the case in which the exterior angle of a triangle is bisected by
a straight line : the demonstration of it is very like to that of the
first case, and upon this account may, probably, have been left out,
as also the enunciation, by some unskilful editor : at least, it is ^r-
tain, that Pappas makes use of this case as an elementary proposi-
tion, without a demonstration of it, in prop. 39 of his 7th book of
Mathematical Collections.

PROP. VIL B. VI.

To this a case is added which occurs not unfirequently in de-
monstration.

PROP. VHL B. VL

It seems plain that some editor has changed the demonstration
that Euclid gave of this proposition: for, after he has demon-
strated, that the triangles are equiangular to one another, he par-
ticularly shows that their sides about the equal angles are pro-

* See page 336.

BOOK VI. . N0TB8. 361

poitlonals, as if this had not been done in the demonstration of
the 4th prop, c^ this boc^; ^is superfluous part is not found ia
the translation from the Arabic, and is now left out.

PROP. IX. B.VL

This is denK>nstrated in a particular case, viz. that in which
the third part of a straight line is required to be cut off; which is
not at all like Euclid's manner: besides, the author of the demon-
8tration« from four magnitudes being proportionals, concludes
that the third of them is the same multiple of the fourth, which
the first is of the second : now, this is no where demonstrated in
the 5th book, .as we now have it : but the editor assumes it from
the confused notion which the vulgar have of proportionals: on
this account, it was necessary to give a general and legitimate de-
monstration <^ this proposition.

PROP. xvni. B. VL

The demonstration of this seems to be vitiated : for the propo-
sition is demonstrated only in the case of quadrilateral figures,
without mentioning how it may be extended to figures of five or
moare sides: besides, from two triangles being equiangular* it is
inferred that a side of the one is to the homologous side of the
other, as another side of the first is to the side homologous to it
of the other, without permutation of the proportionals ; which is
contrary to Euclid's manner, as is clear from the next proposi-
tion; and the same fault occurs again in the conclusion, where
the sides about the equal angles are not shown to be proportion-
als, by reason of again neglecting permutation. On these ac-
counts, a demonstration is given in Euclid's manner, like to that
he makes use of in the 20th prop, of this book: and it is extended
to five-sided figures, by which it may be seen how to extend it to
figufes of any number of sides.

PROP. XXm. B. VI.

Nothing is usually reckoned more difficult in the Elements of
geometry by learners, than the doctrine of compound ratio, which
Theon has rendered absurd and ungeometrical, by substituting
the 5th definition of the 6th book in place of the right definition*
whieh without doabt Eudoxus or Euclid gave, in its proper place*
after the definition of triplicate ratio, &.c, in the 5th book. The-
on's definition is this: a ratio is said to be compounded of ratios
orav m twv Xa^'uv ^riKtxoirriirsi S(p' aouro^ qfoXXa^rXao'ifluf^stf'ai mn<t)(ft riva:
wl»ch Commandine thus translates; '*quando rationem quanti-
tates inter se multiplicata^ aliqvam efficiunt rationem;" that is,
when the quantities of the ratios being multiplied by one another
make a certain ratio. Dr. Wallis translates the word tfijXmonQrss
^ratiooem exponentes," the exponents of the ratios: and Dr. Gr^
gory renders the last words of the definition by ^^illius &cit quan-
titatem**' makes the quantity of that ratio; but in whatever sense
the ''quantities," or ''exponents of the ratios," and their "multi-
plication" be taken, the definition will be ungeometrical and us^
less : for there can be no multiplication but by a number. Now

262 NOTBS. BOOK TI.

the quantity or exponent c^ a ratio (according as Eutochius in
his Comment, on prop. 4, book 2, of Arch, de Sph. et CyL and
the moderns explain that term) is the number which multiplied
into the consequent term of a ratio produces the antecedent, or
which is the same thing, the number which arises by dividing the
antecedent by the consequent; but there are many ratios such,
that no number can arise from the division of the antecedent by
the consequent; ex. gr. the ratio which the diameter of a square
has to the side of it; and the ratio which the circumference of a
circle has to its diameter, and such like. Besides, that there is
not the least mention made of this definition in the writings of
Euclid, Archimedes, Apollonius, or other ancients, though they
frequently make use of compound ratio ; and in this 23d prop, of
the 6th book, where compound ratio is first mentioned, there is
not one word which can relate to this definition, though here, if
in any place, it was necessary to be brought in ; but the right de-
finition is expressly cited in these words : " But the ratio of K to
M is compounded of the ratio of K to L, and of the ratio of L to
M." This definition therefore of Theon is quite useless and ab-
surd : for that Theon brought it into the Elements can scarce be
doubted ; as it is to be found in his commentary upon Ptolemy's
MsyaXTi 2uvTa|i^, page 62, where he also gives a childish explica-
tion of it, as agreeing only to such ratios as can be expressed by
numbers ; and from this place the definition and explication have
been exactly copied and prefixed to the definitions of the 6th book,
as appears from Hervagius's edition: but Zambertus and Com-
mandine, in their Latin translations, subjoin the same to these de-
finitions. Neither Campanus, nor, as it seems, the Arabic manu-
scripts, from which he made his translation, have this definition.
Clavius, in his observations upon it, rightly judges, that the defi-
nition of compound ratio might have been made afi;er the same
manner in which the definitions of duplicate and triplicate ratio
are given ; viz. " That as in several magnitudes that are contuiual
proportionals, Euclid named the ratio of the first to the third, the
duplicate ratio of the first to the second, and the ratio of the first
to the fourth,* the triplicate ratio of the first to the second, that is,
the ratio compounded of two or three intermediate ratios that are
equal to one another, and so on ; so, in like manner, if there be
several magnitudes of the same kind, following one another,
which are not continual proportionals, the first is said to have to
the last the ratio compounded of all the intermediate ratios — only
for this reason, that these intermediate ratios are interposed be-
twixt the two extremes, viz. the first and last magnitudes;, even
as, in the 10th definition of the 5th book, the ratio of the first to
the third was called the duplicate ratio, merely upon account of
two ratios being interposed betwixt the extremes, that are equal
to one another : so that there is no difference betwixt this com*
pounding of ratios, and the duplication or triplication of them
which are defined in the 5th book, but that in the duplication, tri-
plication, &c. of ratios, all the interposed ratios are equal to one
another; whereas, in the compounding of ratios, it is not neces-
sary that the intermediate ratios should -be equal to one another."

BOOK VI. MOTES. 26d

Alao Mr. Edmund Scarburgh, in his English translation of the first
six books, page 238, 266, expressly affirms, that the 5th definition of
the 6th book is supposititious, and that the true definition of ccHn-
pound ratio is contained in the 10th definition of the 5th book, viz.
the definition of duplicate ratio, or to be understood from it, to wit,
in the same manner as Clavius has explained it in the preceding ci-
tation. Yet these, and the rest of the moderns, do notwithstanding
retain this 5th def. of the 6th book, and illustrate and explain it by
long commentaries, when they ought rather to have taken it quite
away from the Elements.

For, by comparing def 5, book 6, with prop. 5, book 8, it will
clearly appear that this definition has been put into the Elements in
place of the right one, which has been taken out of them : because,
in prop. 5, book 8, it is demonstrated that the plane number of
which the sides are C, D has to the plane number of which the
sides are E, Z (see Hervagius's or Gregory's edition,) the ratia
which is compounded of the ratios of their sides ; that is, of the
ratios of C to E, and D to Z : and, by def. 5, book 6, and the expli-
cation given of it by all the commentators, the ratio which i& com*
pounded of the ratios of C to E, and D to Z, is the ratio of the pro-
duct made by the multiplication of the antecedents, C, D, to the pro-
duct by the consequents E, Z, that is, the ratio of the plane number
of which the sides are C, D to the plane number of which the sides
are K, Z. Wherefore the proposition which is the 5th def. of book
6, is the very same with the 6th prop, of book 8, and therefore it
ought necessarily to be cancelled in one of these places ; because it
is absurd that the same proposition should stand as a definition in
one place of the Elements, and be demonstrated in another place of
them. Now, there is no doubt that prop. 5, book 8, should have a
place in the Elements, as the same thing is demonstrated in it con-
cerning plane numbers, which is demonstrated in prop. 23, book 6,
of equiangular parallelograms ; wherefore def 5, book 6, ought not
to be in the Elements. And from this it is evident that this defini-
tion is not Euclid's but Theon's, or some other unskilful geometer's.

But nobody, as far as I know, has hitherto shown the true use of
compound ratio, or for what purpose it has been introduced into
geometry : for every proposition in which compound ratio is made
use of, may without it be both enunciated and demonstrated. Now
the use of compound ratio consists wholly in this, that by means of
it circumlocutions may be avoided, and thereby propositions may be
more briefly either enunciated or demonstrated, or both may be
done : for instance, if this 23d proposition of the sixth book were to
be enunciated, without mentioning compound ratio, it might be done
as follows. If two parallelograms be equiangular, and if as a side
of the first to a side of the second, so any assumed straight line be
made to a second straight line: and as the other side of the first to
the other side of the second, so the second straight line be made a
third. The first parallelogram is to the second, as the first straight
line to the third. And the demonstration would be exactly the same
as we now have it. But the ancient geom^ers, when they observed
this enunciation could be made shorter, by giving a name to the

204 SOTfiS. BOOK vx.

ratio which the first straight line has to the last, by which name
the intermediate ratios might likewise be signified, c^ the. first to
the second, and of the second to the third, and so on, if there were
more of them, they called this ratio of the first to the last the ratio
compounded of the ratios of the first to the second, and of the
second to the third straight line : that is, in the present example, of
the ratios whlcJi are the same with the ratios of the sides, aiid by
this they expressed the proposition more briefly, thus : if there be
two equiangular parallelc^ams, they have to one another the ratio
wi^ich is the same with that which is compounded of ratios that
are the same with the ratios of the sides. Which is shorter than
the preceding enunciation, but has precisely the same meaning.
Or yet shorter thus: equiangular parallelograms have to one
another the ratio which is the same with that which is com*
pounded of the ratios of their sides. And these two enunds^onsy
the first especially, agree to the demonstration which is now in the
Greek. The proposition may be more briefiy demonstrs^ed, as
Oandalia does, thus: let ABCD, CEFG, be two equiangular parallelo-
grams, and complete the parallelogram CDHG, then, because there
are three parallelograms AC, CH, CF, the first AC (by the definlticm
of compound ratio) has to the third CF, the ratio which is oom-
pounded of the ratio of the first AC to the AD n

second CH, and of the ratio of CH to the
third CF ; but the parallelogram AC is to the
parallelogram CH, as the straight line EC to

CG ; and the parallelogram CH is to CF, as q

the straight line CD is to CE: therefore the " '^

parallelogram AC has to CF the ratio which
is compounded of ratios that are the same
with the ratios of the sides. And to this de- ^ F

monstration agrees the enunciation which is
at present in the text, viz. equiangular parallelograms have to one
another the ratio which is compounded o( the ratios of the sides;
for the vulgar reading, ** which Is compounded of their sides,'* is
absurd. But in this edition, we have kept the demonstration which
is in the Greek text, though not so short as Candalla's ; because the
way of finding the ratio which is compounded of the ratios of the
sides, that is, of finding the ratio of the parallelograms, is shown in
that, but not in Candalla's demonstration ; whereby beginners may
learn, in like cases, how to find the ratio which is compounded ai
two or more given ratios*

From what has been said, it may be observed, that in any mag-
nitudes whatever of the same kind A, B, C, D, &c. the ratio conk-
pounded df the ratios of the first to the second, of the second to
the third, and so on to the last, is only a jiame or expression, by
which the ratio which the first A has to the last D is signified, and
by which at the same time the ratios of all the magnitudes A
to B, B to C, C to D, from the first to the last, to one another ;
whether they be the same, or be not the same, are indicated ; as in
magnitudes which are continual proportionals A, B, C, D, &c the
duplicate ratio of the first to the second is only a name or eac<*

;

B C

■

BOOK rt NOTES. %65

pr^ssioh by which the ratio of the first A to the third C is signified*
and by which, at the same time, is shown that there are two ratios
of the magnitudes, from the first to the last, viz. of the first A to the
second B, and of the second B to the third or last C, which are the
same with one another ; and the triplicate ratio of the first to the
second is a name or expression by which the ratio of the first A to
the fourth D is signified, and by which, at the same time, is shown
that there are three ratios of the magnitudes, firom the first to the
last, viz. of the first A to the second B, and of B to the third C, and
of C to the fourth or last D, which are all the same with one an-
other^ and so in the case of any other multiplicate ratios. And that
this is the right explication of the meaning of these ratios is plain
from the definitions of duplicate and triplicate ratio, in which Euclid
makes use of the word 'ksysrou, is said to be, or is called, ^hich
word, he, no doubt, made use of also in the definition of compound
ratio, which Theon, of some other, has expunged from tlie Elements ;
for the very same word is still retained in the wrong definition ci
compound ratio, which is now the 5th of the 6th book : but hi the
citation of these definitions it is sometimes refined, as in the de-
monstration of prop. 19, book 6, "the first is i&aid to have, s^fi*
Xtyerai, to the third the duplicate ratio," &c. which is wrong trans-
lated by Commandine and others, " has" instead of *• is said to
have :" and sometimes it is left out, ai^ in the demonstration of prop.
33, of the eleventh book, in which we find *' the first has, sp^«i, to the
third the triplicate ratio;" but without doubt sxeu "has" in this
place signifies the same as sp^siv "key&rai, is said to have: so likewise
in prop. 23, B. 6, \*e find this citation, " but the ratio of K t6 M is
compounded, (fvyxstrat of the ratio of K to L, and the ratio of L to
M," which is a shorter way of expressing the same thing, which, a<N
cording to the definition, ought to have been expressed by (fvyxsuf^tts
"ksyBttnf, is said to be compounded.

From these remarks, together with the proposition subjoined, to
the 5th book, all that is found concerning compound ratio, either in
the ancient or modern geometers, may be understood and explained.

PROP. XXIV. R VI.

It seems that some unskilful editor has made up this demonstra-
tion as we now have it, out of two others ; one of which may be
made from the 2d prop, and the other from the 4th of this book : for
after he has, from the 2d of this book, and composition and permuta-
tion, demonstrated, that the sides about the angle common to the
two parallelograms are proportionals, he might have immediately
concluded, (hat the sides about the other equal angles were propor-
tionals, viz. firom prop. 34, B. 1 , and prop. 7, book 6. "fhis he does
not, but proceeds to show, that the triangles and parallelograms are
equiangular ; and in a tedious way by help of prop. 4, of this book,
and the 22d of book 5, deduces the same conclusion : from which it
is plain that this ill composed demonstration is not Enclid*s : these
superfluous things tre now feft out, and a more simple demonstra!-
tion is gfven fi^om the fourth prop, of this book, the same which is

'34

3Qf^ NOTES. BOOK VI.

in the trapslation from the Arabic, by lielp of the 2d prop, and com-
position : but in this the author neglects permutation, and does not
show the parallelograms to be equiangular, as is proper to do for the
sake of b^^inners.

PROP. XXV. B. VI.

It is very evident that the demonstration which Euclid had givea
of this proposition has been vitiated by some unskilful hand : for,
after this editor had demonstrated that " as the rectilineal figure ABC
is to the rectilineal KGH, so is the parallelogram BE to the parallelo-
gram EF ;'* nothing more should have been added but this, " and
the rectilineal figure ABC is equal to the parallelogram BE : therefore
the rectilineal KGH is equal to the parallelogram EF," viz. from
prop. 14, book 5. But betwixt these two sentences he has inserted
this ; " wherefore, by permutation, as the rectilineal figure ABC to the
parallelogram BE» so is the rectilineal KGH to the parallelogram EF;"
by which it is plain, he thought it was not evident to conclude, that
the second of four proportionals is equal to the fourth from the
equality of the first and third, which is a thing demonstrated in the
14th prop, of B. 5, as to conclude that the third is equal to the fourth,
from the equality of the first and second, which is no where demon-
strated in the elements as we now have them : but though this pro-
position, viz. the third of four proportionals is equal to the fourth, if
the first be equal to the second, had been given in the Elements by
Euclid, as very probably it was, yet he would not have made use of
it in this place ; because, as was said, the conclusion could have
been immediately deduced without this superfluous step, by permu-
tation : this we have shown at the greater length, both because it
affords a certain proof of the vitiation of the text of Euclid ; for the
very same blunder is found twice in the Greek text of prop. 23,
book 11, and twice in prop. 2, B. 12, and in the 5, 11, 12, and 18th
of that book ; in which places of book 12, except the last of them,,
it is rightly left out in the Oxford edition of Commandine's transla-
tion ; and also that geometers may beware of making use of permu-
tation in the like cases : for the moderns not unfrequently commit
this mistake, and among others Commandine himself, in his commen-
tary on prop. 6, book 3, p. 6, b. of Pappus Alexandrinus, and in
other places : the vulgar notion of proportionals has, it seems, pre-
occupied many so much, that they do not sufficiently understand the
true nature of them.

Besides, though the rectilineal figure ABC, to which another is to
be made similar, may be of any kind whatever ; yet in the demon-
stration the Greek text has " triangle" instead of " rectilineal figure,"
which error is corrected in the above named Oxford edition.

PROP. XXVIL B. VI.

The second case of this has aXku^^ otherwise, prefixed to it, aa
if it was a different demonstration, which probably has been done
by some unskilful librarian. Dr. Gregory has rightly left it out :

BOOK Vr. H0TB8. tf7

the scheme of this second case ''ought to be marked with the same
letters of the alphabet which are in the scheme of the first, as is now
done.

PROP. XXVin. and XXIX. B. VI.

These two problems, to the first of which the 27th prop, is neces-
sary, are the most general and useful of all in the Elements, and are
most frequently made use of by the ancient geometers in the solution
of other problems ; and therefore are very ignorantly left out by
Tacquet and Dechales in their editions of the Elements, who pretend
that they are scarce of any use. The cases of these problems, where-
in it is required to apply a rectangle which shall be equal to a given
square, to a given straight line, either deficient or exceeding by a
square ; as also to apply a rectangle which shall be equal to another
given, to a given straight line, deficient or exceeding by a square,
are very often made use of by geometers. And, on this account, it
is thought proper, for the sake of beginners, to give their construc-
tions as follows :

1. To apply a rectangle which shall be equal to a given square, to
a given straight line, deficient by a square ; but the given square must
not be greater than that upon the half of the given line.

Let AB be the given straight line, and let the square upon the
given straight line C be that to which the rectangle to be applied
must be equal, and this square, by the determination, is not greater
than that upon half of the straight line AB.

Bisect AB in D, and if the square upon AD be eqc(al to the square
upon C, the thing required is done : but if it be not equal to it, AD
must be greater than C, according to H

the determination ; draw DE at right L
angles to AB, and make it equal to
C : produce ED to P, so that EF be A
equal to AD or DB, and frorai^the
centre E, at the distance EF, de-
scribe a circle meeting AB in G, and
upon GB describe the square GBKH,
and complete the rectangle AGHL ;
also join EG. And because AB is
bisected in D, the rectangle AG, GB together with the square of DG
is equal (5. 2.) to (the square of DB, that is, of EF or EG, that is, to)
the squares of ED, DG : take away the square of DG from each of
these equals ; therefore the remaining rectangle AG, GB is equal to
the square of ED, that is, of C ; but the rectangle AG, GB is the rect-
angle AH, because GH is equal to GB ; therefore the rectangle AH is
equal to the given square upon the straight line C. Wherefore the
rectangle AH, equal to the given square upon C, has been applied to
the given straight line AB deficient by the sqiSare GK. Which was
to be done.

2. To apply a rectangle which shall be equal to a given square, to
a given straight line, exceeding by a square.

X0TB8.

Boaic VI.

Let AB be the given straight line, and let the square upcm the
given straight line C be that to which the rectangle to be applied
must be equal.

Bisect AB in D, and draw BE at right angles to it, so that BE be
equal to C ; and having joined DE, from the centre D at the distance
DE describe a circle meeting AB produced in G; upon BG describe
th« square BGHK, and complete the
rectangle AGHL. And because AB is
biseeted in D, and produced to G, the
rectangle AG, GB together with the
square of DB is equal (6. 2.) to (the
square of DG, or DE, that is, to) the
squaresof EB, BD. From each of these p AD B G

equals take the square of DB; therefore,

the remaining rectangle AG, GB is equal C

to the square of BE, that is, to the square upon C. But the rectan-
gle AG, GB is the rectangle AH, because GH is equal to GB : there-
fore the rectangle AH is equal to the square upon C. Wherefore the
rectangle AH, equal to the given square upon C, has been applied to
the given straight line AB, exceeding by the square GE. Which was
to be done.

3. To apply a rectangle to a given straight line which shall be
equal to a given rectangle, and be deficient by a square. But the
given rectangle must not be greater than the square upon the half

of the given straight line.

Let AB be the given straight line, and let the given rectangle be
that which is contained by the straight lines C, D which is not great*
er than the square upon the half of AB ; it is required to apply to AB
a rectangle equal to the rectangle 0, D, deficient by a square.

Draw AE, BF at right angles to AB, upon the same side of it, and
make AE equal to C, and BF to D : join EF, and bisect it in G ; and
from the centre G, at the distance Qj^ describe a circle meeting AE
again in H : join HF, and draw GK parallel to it, and GL parallel to
AE, meeting AB in L.

Because the angle EHF in a semicircle is equal to the right
angle EAB, AB and HF are parallels, and AH and BF are paral-
lels; wherefore AH is equal to BF, and the rectangle EA, AH
equal to the rectangle EA, BF, that is to the rectangle O, D : and
because EG, GF are equal to one another, and AE, LG, BF pa-
rallels; therefore AL and LB are equal: also EK is equal to KH
(3. 3.), and the rectangle C, D, from the determination, is not
greater thAa the square of AL, the half of AB ; wherefore the rect-
angle BA, AH is not greater than the square of AL, that is of
KG : add to each the square of KE ; therefore the square (6. 2.)
of AK is not greater than the squares of EK, KG, that is, than

BOOK VI.

MOTM.

the square of EG; and i^onse-
quently the straight line AK or
GL is not greater than GE. £
Now, if GE be equal to GL, the
circle EHF touches AB in L,
and therefore the square of AL
is (36. 3.) equal to the rectangle
EA, AH, that is, to the given
rectangle C, D ; and that which
was required is done: but if
EG, GL be unequal, EG must
be the greater : and therefore the
circle EHF cuts the straight
Hne AB: let it cut it in the
points M, N, and upon NB describe the square NBOP, and com-
plete the rectangle ANPQ: because LM is equal to (3. 3.) LN,
and it has been proved that AL is equal to LB ; therefore AM is
equal to NB, and the rectangle AN, NB equal to the rectangle
NA, AM, that is, to the rectangle (Cor. 36. 3.) EA, AH, or the
rectangle C, D: but the rectangle AN, NB is the rectangle AP,
because PN is equal to NB: therefore the rectangle AP is equal
to the rectangle C, D ; and the rectangie AP equal tp the given
rectangle C, D has been applied to the given straight line AB» de-
ficient by the square BP. Which was to be done.

4. To apply a rectangle to a given straight line that shall be
equal to a given rectangle, exceeding by a square.

Let AB be the given straight line, and the rectangle C, D th^
given rectangle, it is required to apply a rectangle to AB equal to
CD, exceeding by a square.

Draw AE, BP»at right angles to AB, on the contrary sides* of
it, and make AE equal to C, and BF equal to D : join EF, and
bisect it in G; and from the centre G, at the distance GE, de-
scribe a circle meeting AE again in H ; join HF, ,and draw CL
parallel to AE; let the circle
meet AB produced in M, N,
and upon BN describe the
square NBQP, and complete
the rectangle ANPCl; because
the angle EHF in 9, semicircle
is equal to the right angle
EAB, AB and HF are paral-
lels, and therefore AH and BF
are equal, and the rectangle
EA, AH equ^l to the rectangle
EA, BF, that is, to the rect-
angle C, D: and because ML

is equal to LN, and AL to LB, therefore MA is equal to BN, and
the rectangle AN, NB to MA, AN, that is (36. 3.) to the rectangle
EA, AH, or the rectangle C, D : therefore the rectangle AN, NB,
that is, AP, is equal to the rectangle C, D; and to the given
straight line AB the rectangle AP has beeif applied equal to the

270 NOTES. BOOK Vt.

given rectangle C, D, exceeding by the square BP. Which was to
i3e done.

Willebrprdus Snellius was the first, as far as I know, who gave
these constructions of the 3d and 4th problems, in- his Apollonius
Batavius ; and afterwards the learned Dr. Halley gave them in the
scholium of the 18th prop, of the 8th book of Apoilonius*s conies re-
stored by him.

The 3d problem is otherwise enunciated thus: to cut a given
straight line AB in the point N, so as to make the rectangle AN, NB
equal to a given space ; or, which is the same thing, having given AB
the sum of the sides of a rectangle, and the magnitude of it being
likewise given, to find its sides.

And the fourth problem is the same with this. To find the point
N in the given straight line AB produced so as to make the rectangle
AN, NB equal to a given space : or, which is the same thing, having
given AB the difference of the sides of a rectangle, and the magnitude
of it, to find the sides. •

PROP. XXXI. RVI.

In the demonstration of this, the inversion of proportionals is twice
neglected, apd is now added, that the conclusion may be legitimately
made by help of the 24th prop, of B. 5. as Clavius had done.

PROP. XXXn. B. VI.

The enunciation of the preceding 26th prop, is not general enough ;
because not only two similar parallelograms that have an angle com-
mon to both, are about the same diameter; but likewise two similar
parallelograms that have vertically opposite angles, have their diame-
ters in the same straight line : but there seems to Have been another,
and that a direct demonstration of these cases, to which this 32d pro-
position was needful : and the 32d may be otherwise and something
more briefly demonstrated as follows.

PROP. XXXII. B. VI.

If two triangles which have two sides of the one, &c.

Let GAF, HFC be two triangles which have two sides AG, OF pro-
portional to the two sides FH, HC, viz. AG to GF, as FH to HC ; and
let AG be parallel to FH, and GF to HC; A G D

AF and FC are in a straight line.

Draw CK parallel (31. 1.) to FH, and let
it meet GF produced in K : because AG, E } N j ^ ■■ ■ | H
KC are each of them parallel to FH, they
are parallel (30. 1.) to one another, and
therefore the alternate angles AGF, FKC

are equal : and AG is to GF, as (FH to HC, ry ir r*

that is 34. 1.) CL to KF ; wherefore the tri- ^ ^ ^

angles AGF, CKF are equiangular, (6. 6.) and the angle AFG equal
to the angle CFK : but GFK is a straight line, therefore AF and FC
are in a straight line (14. 1.).

BQOK XI. NOTIS. 271

The 26th prop, is demonstrated from the 32d, as follows :

If two similar and similarly placed parallelograms have an angle
common to both, or vertically opposite angles ; their diameters are
in the same straight line.

First, let the parallelograms ABCD, AEFG have the angle BAD
common to both, and be similar and similarly placed ; ABCD, AEFG
are about the same diameter.

Produce EF, GF, to H, K, and join FA, FC : then because the
parallelograms ABCD, AEFG are similar, DA is to AB, as GA to
AE : wherefore the remainder DG is (Cor. AG D

19. 5.) to the remainder EB, as GA to
AE : but DG is equal to FH, EB to HC, ,

and AE to GF : therefore as FH to HC, E | "*K 1 H

so is AG to GF ; and FH, HC are parallel
• to AG, GF ; and the triangles AGF, FHC
are joined at one angle in the point F:
wherefore AF,tFC are in the same straight h"
line (32. 6.) a j^ v.

Next, let the parallelograms KHFC, GFEA, which are similar and
similarly placed, have their angles KFH, GFE vertically opposite ;
their diameters AF, FC are in the same straight line.

Because AG, GF are parallel to FH, HC ; and that AG is to GF,.
as FH to HC; therefore AF, FC are in the same straight line
(32. 6.).

PROP, xxxra. B. VI.

The words " because they are at the centre," are left out, as the
addition of some unskilful hand.

In the Greek, as also in the Latin translation, the words a sruys
" any whatever," are left out in the demonstration of both parts of
the proposition, and are now added as quite necessary ; and in the
demonstration of the second part, where the triangle BGC is proved
to be equal to CGK, the illative particle aja in the Greek text ought
to be omitted.

The second part of the proposition is an addition of Theon's, as
he tells us in his commentary on Ptolemy's MsyaXrj 2uv«ragis, p. 50.

PROP. B. C. D. B. VI.

These three propositions are added, because they are frequently
made use of by geometers.

DEF. IX. and XL B. XI.

The similitude of plane figures is defined from the equality of
their angles, and the proportionality of the sides about the equal
angles; for from the proportionality of the sides only, or only
from the equality of the angles, the similitude of the figures does
not follow, except in the case when the figures are triangles : the
similar position of the sides which contain the figures, to one ano-
ther, depending partly upon each of these : and, for the same rea-
son, those are similar solid figures which have all their solid angles
equal, each to each, and are contained by the same number of

in NOTES.' BOOK XI.

similar plane figures : for there are some solid figures contained by
similar plane figures, of the same number, and ev^n of the same
magnitude, that are neither similar not equal, as shall be demon-
strated after the notes on the 10th definition; upon this account it
was necessary to amend the definition of similar solid figures, and
to place the definition of a solid angle before it : and from this and
the 10th definition, it is sufficiently plain how much the Elements
have been spoiled by unskilful editors.

DEP. X. B. XL

Since the meaning of the word '< equal" is known and established
before it comes to be used in this definition ; therefore the proposition
which is the 10th definition of this book, is a theorem, the truth or
fiilsehood of which ought to be demonstrated, not assumed ; so that*
Theon, or some other editor, has ignorantly turned a theorem wiiich
ought to be demonstrated into this 10th definition ; ttiat figures are
similar, ought to be proved from the definition of similar figures;
that they are equal, ought to be. demonstrated from the axiom^
** Magnitudes that wholly coincide, are equal to one another :" or
from prop. A, of book 5, or the 9th prop, or the 14th of the same
bookf from one of which the equality of all kind of figures must ulti-
ixmtely be deduced. In the preceding books, Euclid has given no
definition of equal figures, and it i§ certain he did not give this : for
what is caUed the 1st def. of the 3d book is really a theorem in
which these circles are said to be equal, that have the straight lines
from their centres to the circumferences equal, which is plain, from
the definition of a circle ; and therefore has by some editor been
improperly placed among the definitions. The equality of figures
ought not to be defined, but demonstrated: therefore, though it were
true, that solid figures contained by the same number of similar
and equal plane figures are equal to one another, yet he would justly
deserve to be blamed who would make a definition of this proposi-
tion, which ought to be demonstrated. But if this proposition be not
true, must it not be confessed, that geometers have, for these thirteen
hundred years, been mistaken in this elementary matter 1 And this
should teach us modesty, and to acknowledge how little, through the
weakness of our minds, we are able to prevent mistakes, even in the
principles of sciences which are justly reckoned amongst the most
certain ; for that the proposition is not universally true, can be shown
by many examples ; the following is sufficient.

Let there be any plane rectilineal figure, as the triangle ABC,
and from a point within it draw (12r 11.) the straight line DE
at right angles to the plane ABC; in DE take DE, DF equal to
one another, upon the opposite sides of the plane, and let G be
any point in EF ; join DA, DB, DC ; EA, EB, EC ; FA, FB, FC ;
GA, GB, GC; because the straight line EDF is at right angles
to the plane ABC, it makes right angles with DA, DB, DC which
it meets in that plane : and in the triangles EBD, FBD, ED and
DB are equal to FD and DB, each to each, and they contain right

BOOK 3EI.

NOTES.

370

angles; thwefinre tiie base EB G

is equal (4. 1.) to the base FB;
in the same manner £14 is equal
to FA, and &C to FC: and in
the triangles EBA, FBA, EB,
BA are equal to FB, BA; and
the base BIA, is equal to the base
FA ; wherefore the angle EBA
is equal (8. 1 .) to the angle FBA,
and the triangle EIBA equal (4.
I.) to the triangle FBA, and the
other angles equal to the other
angled; therefore these triangles
are similar (4. 6. 1. de£) : in the
same manner the triangle EBC
is similar to the triangle FBC, F

and the triangle EAC to FAC ; therefore there are two solid figures^
each of which is contained by six triangles, one of them by three tri-
angles, the common vertex of which is the point G, and thear bases
the straight lines AB, BC, CA, and by three other triangles the coni-
mon vertex of which is the point E, and their bases the same lines
AB, BG, GA : the other solid is contained by the same three triangles
the common vertex of which is G, and their bases AB, BC, CA ; and
by three other triangles of which the common vertex is the point F,
and their bases the same straight lines AB, BC, CA : now the three
triangles GAB, GBC, GCA are common to both solids, and the three
others EAB, EBC, EGA of the first solid have been shown equal and
similar to the three others FAB, FBC, FCA of the other solid, each
to each; therefore these two solids are contained by the same num-^
ber of equal and similar planes : but that they are not equal is mani- *
fest, because the first of them is contained in the other : therefore it is
not universally true that solids are equal which are contained by the
same number of equal and similar planes.

CoR. From this it appears that two unequal solid angles may be
contained by the same number of equal plane angles.

' For the solid angle at B, which is contained by the four plane an-
gles EiBA, EBC, GBA, GBC is not equal to the solid angle at the
same point B, which is contained by the four plane angles FBA, FBC,
GBA, GBC ; for this last contains the other : and each of them is con-
tained by four plane angles which are equal to one another, each to
each, or are the self same; as has been proved: and indeed there
may be innumerable solid angles all unequal to one another, which
are each of them contained by plane an^es that are equal to one
another, each to each : it is likewise manifest that the before men-
tioned solids are not similar, since their solid angles are not all
equal.

And that there may be innumerable solid angles all unequal to one
another, which are each of them contained by the same plane angles
disposed in the same order, will be plain from the three following
propositions.

274 NOTES. BOOK

PROP, I. PROBLEM.

Three magnitudes, A, B, C being given, to find a fourth such, that
every three shall be greater than the remaining one.

Let D be the fourth : therefore D must be less than A, B, C to-
gether : of the three, A, B, C, let A be that which is not less thai^
either of the two B and C : and first, let B and C together be not
less than A : therefore B, C, D together are greater than A ; and be-
cause A is not less than B ; A, C, D together are greater than B : ia
the like manner. A, B, D together are greater than C : wherefore,
in the case in which B and C together are not less than A, any
magnitude D which is less than A, B, C together, will answer the
problem.

But if B and C together be less than A ; then, because it is required
that B, C, D together be greater than A, from each of these -taking
away B, C, the remaining one D must be greater than the excess of
A above B and C ; take therefore any magnitude D which is less than

A, B, C, together, but greater than the excess of A above B and C :
then B, C, D together are greater than A ; and because A is greater
than either B or C, much more will A and D together with either of
the two B, C be greater than the other ; and by the construction, A»

B, C are together greater than D.

Cor. If besides it be required, that A and B together shall not be
less than C and D together ; the excess of A and B together above
C must not be less than D, that is, D must not be greater than that
excess.

PROP. n. PROBLEM.

Four magnitudes A, B, C> D being given, of which A and B to-
gether are not less than C and D together, and such that any tliree
of them whatever are greater than the fourth ; it is required to find
a fifth magnitude E such, that any two of the three A, B, E shall
be greater than the third, and also that any two of the three C, D,.
E shall be greater than the third. Let A be not less than B^ and C
not less than D.

First, let the excess of C above D be not less than the excess of
A above B : it is plain that a magnitude E can be taken which is less
than the sum of C and D, but greater than the excess of C above D ;
let it be taken ; then E is greater likewise than the excess of A above
B : wherefore E and B together are greater than A ; and A is not less
than B ; therefore A and E together are greater than B : and, by the
hypothesis, A and B together are not less than C and D together, and
C and D together are greater than E ; therefore likewise A and B are
greater than E.

But lot the excess of A above B be greater than the excess of
C above D ; and because, by the hypothesis, the three B, C, D are
together greater than the fourth A : C and D together are greater
than the excess of A above B: therefore a magnitude may be
taken which is less than C and D together, but greater than the

BOOK XI.

NOTES.

276

excess of A above B. Let this magnitude be E ; and because E Is
greater than the excess of A above B, B together with E is gfeatd-
than A : and, as in the preceding case, it may be shown that A to-
gether with E is. greater than B, and that A together with B is
greater than E : therefore, in each of the cases, it has been shown
that any two of the three A, B, E are greater than the third.

And because in each of the cases E is greater than the excess of
C above D, E together with D is greater than C ; and, by the hypo-
thesis, C is not less than D ; therefore E togethtr with C is greater
than D; and, by the construction, C and D together are greater than
E : therefore any two of the three, C, D, E are greater than the
third. ^

PROP. ra. THEOREM.

There may be innumerable solid angles all unequal to one another,
each of which is contained by the same four plane angles, placed in
the same order.

Take three plane angles A, B, C, of which A is not less than
either of the other two, and such, that A and B together are less
than two right angles : and by problem 1, and its corollary, find a
fourth angle D such, that any three whatever of the angles A, B, C,
D be greater than the remaining angle, and such, that A and B to-
gether be not less than C and D together : and by problem 2, find
a fifth angle E such, that any two of the angles A, B, E be greater
than the third, and also that any two of the angles C, D, E, be great-

A E C F

er than the third : and because A and B together are less than two
right angles, the double of A and B together is less than four right
angles : but A and B together are greater than the angle E ; where-
fore the double of A, B together is greater than the three angles A,

B, E together, which three are consequently less than four right an-
gles ; and every two of the same angles A, B, E are greater than
the third; therefore, by prop. 23. 11, a solid angle maybe made
contained by three plane angles equal to the angles A, B, E, each to
each. Let this be the angle F contained by the three plane angles
GFH, HFK, GFK which are equal to the angles A, B, E, each to
each : and because the angles C, D together are not greater than
the angles A, B together, therefore the angles C, D, E are not great-
er than the angles A, B, E : but these last three are less than four
right angles, as has been demonstrated : wherefore also the angles

C, D, E are together less than four right angles, and every two of

276

NOTBB.

BooiL m.

tbefn are gfe^t^ than the third ; therefore a solkl angle nia|r be
made which shall be contained by three plane angles equal to the
angles G, D, £, each to each (23. 11.): and by prop. 26. 11, at the
point F in the straight line FG a solid angle may be made eqiial to

that which is contained by the three plane angles that are equal to
the angles C, D, E : let this be made, and let the angle GFK, which
Is equal to E, be one of the three ; and let KFL, GFL be the other
two which are equal to the angles C, D, each to each. Thus there
is a solid angle constituted at the point F, contained by the four
plane angles GFH, HFK, KFL, GFL which are equal to the angles
A, B, C, D, each to each.

Again, find another angle M such, that every two of the three an*
gles A, B, M be greater than the third, and also every two of the
three C, D, M be greater than the third : and, as in the preceding
part, it may be demonstrated that the three A, B, M are less than
four right angles, as also that the N

three C, D, M are less than four right
angles. Make therefore (23. 11.) a
solid angle at N contained by the
three plane angles ONP, PNCl, ONQ,
which are equal to A, B, M, each to
each: and by prop. 26. 11, make at
the point N in the straight line ON a O
solid angle contained by three plane
angles of which one is the angle ONQ, equal to M, and the other two
are the angles QNR, ONR, which are equal to the angles C, D, each
to each. Thus, at the point N, there is a solid angle contained by
the four plane angles ONE, PNQ, QNR, ONR which are equal to
the angles A, B, C, D, each to each. And that the two solid angles
at the points F, N, each of which is contained by the above named
four plane angles, are not equal to one another, or that they cannot
coincide, will be plain by considering that the angles GFK, ONQ,
that is, the angles E, M, are unequal by the construction; and
therefore the straight luies GP, FK cannot coincide with ON, NQ,
nor consequently can the solid angles, which therefore are unequal.

And because from the four plane angles A, B, C, D, there can
be found innumerable other anglei^ that will serve the same pur<

BOOK Xf. N0TE3. 277

pose with the angles E and M ; it is plain that innumerable other
solid angles may be constitute^ which are each contained by the
same four plane iangles, and ail of them unequal to one another.

a. E. D.

And from this it appears that Clavius and other authors are
mistaken, who assert that those solid angles are equal which are
contained by the same number of plane angles that are equal to
one another each to each. Also it is plain that the 26th prop, of
book 11, is by no means sufficiently demonstrated, because the
equality of two solid angles, whereof each is contained by three
plane angles which are equal to one another, each to each, is only
assumed, and not demonstrated.

PROP. I. B. XI.

The words at the end of this, '* for a straight line cannot meet
a straight line in more than one point,'' are left out, as an addi-
tion by some unskilful hand; for this is to be demonstrated, not
assumed.

Mr. Thomas Simson, in his notes at the end of the 2d edition
of his Elements of Geometry, p. 262^ after repleating the words of
this note, adds, ^Now, can it possibly show any want of skiH in
an editor (he means Euclid or Theon) to refer to an axiom which
Euclid himself hath laid down, book 1, No. 14, (he means Barrow's
Euclid, for it is the 10th in the Greek), and not to have demon-
strated, what no man can demonstrate?" But all that in this case
can follow from that axiom is, that, if two straight lines could
meet each other in two points, the parts of them betwixt these
points must coincide, and so they would have a segment betwixt
these points common to both. Now, as it has not been shown in
Euclid, that they cannot have a common segment, this does not
prove that they cannot meet in two points, from which their not
having a common segment is deduced in the Greek edition: but,'
on the contrary, because they cannot have a common segment, as
is shown in cor. of 11th prop, book 1, of 4to edition, it follows
plainly that they cannot meet in two points, which the remarker
says no man can demonstrate.

Mr. Simson, in the same notes, p. 265, justly observes, that in
the coroDary of prop. 11, book 1* 4to edition, the straight lines
AB, BD, BC are supposed to be all in the same plane, which
cannot be assumed in 1st prop, book 11. This, soon after the 4to
edition was published, I observed, and corrected as it is now in
this edition: he is mistaken in thinking the 10th axiom he men-
tions here to be Euclid's; it is none of Euclid's, but is the 10th in
Dr. Barrow's edition, who had it from Herigon's Cursus, vol. 1,
and in place of it the corollary of 10th prop, book 1, was added.

PROP. n. RXI.

This proposition seems to have been changed and vitiated by
some editor : for all the figures defined in the 1st book of the Ele-
ments, and among them^ triangles, are, by the hypothesis, plane
figures ; that is, such as are described in a plane ; wherefore the
second part of the enunciation needs no demonstration. Besides,

278 NOTES. BOOK XI.

a convex superficies may be terminated by three straight lines
meeting one another; the thing that should have been demon-
strated is, that two or three straight lines, that meet one another,
are in one plane. And as this is not sufficiently done, the enuncia-
tion and demonstration are changed into those now put into the text.

PROP. m. B. XI.

In this proposition the following words near to the end of it are
left out, viz. '* therefore DEB, DFB are not straight lines ; in the
like manner it may be demonstrated that there can be np other
straight line between the points D, B;" because from this that two
lines include a space, it only follows that one of them is not a
straight line : and the force of the argument lies in this, viz. if the
common section of the planes be not a straight line, then two
straight lines could include a space, which is absurd ; therefore
the common section is a straight line.

PROP. IV. RXI.

The words "and the triangle AED to the triangle BEC'* are
omitted, because the whole conclusion of the 4th prop. l)ook I, has
been so often repeated in the preceding books, it was needlesij to re-
peat it here.

PROP. V. B. XL

In this, near to the end sifiieiSu, ought to be left out in the Greek
text : and the word " plane" is rightly left out in the Oxford edition
of Commandine^s translation.

PROP. vn. B. XL

This proposition has been put into this book by some unskilful
editor, as is evident from this, that straight lines which are drawn
from one point to another in a plane, are in the preceding books,
supposed to be in that plane : and if they were not, some demon-
strations in which one straight line is supposed to meet another
would not be conclusive, because these lines would liot meet one
another: for instance, in prop. 30, book 1, the straight line GK
would not meet EF, if GK were not in the plane in which are the
parallels AB, CD, and in which, bv hypothesis, the straight line
EF is ; because, this 7th propositioifts demonstrated by the preced-
ing 3d, in which the very thing which is proposed to be demon-
strated in the 7th, is twice assumed, viz. that the straight Ime drawn
from one px)int to another in a plane, is in that plane ; and the same
thing is assumed in the preceding 6th prop, in which the straight
line which joins the points B, D that are in the plane to which AB
and CD are at right angles, is supposed to be in that plane : and the
7th, of which another demonstration is given, is kept in the boolc
merely to preserve the number of the propositions ; for it is evident
from the 7th and d5th definitions of the Ist book, though it had not
been in the Elements.

PROP. vin. B. XI.

In the Greek, and in Commandine's and Dr. Gregory's transla-
tions, near to the end of this proposition, are the following words ;

BOOK XI. NOTBS. 279

'' but DC is in the plane through BA, AD,'* instead of which, in the
Oxford edition of Commandine's translation, is rightly put ** but DC
is in the plane through BD, DA :'* but all the editiohs have the fol-
lowing words, viz. '* because AB, BD are in the plane through BD,
DA, and DC is in the plane in which are AB, BD," which are mani-
festly corrupted, or have been added to the text; for there was not
the least necessity to go so far about to show that DC is in the same
plane in which are BD, DA, because it immediately follows from
prop» 7, preceding, that BD, DA are in the plane in which are the
parallels AB, CD : therefore, instead of these words, there ought
only to be '* because all three are in the plane in which are the
parallels AB, CD."

PROP. XV. B. XI.

After the words " and because BA is parallel to GH," the follow-
ing are added, ** for each of them is parallel to DE, and are not both
in the same plane with it," as being manifestly forgotten to be put
into the text.

PROP. XVI. B. XL

In this near to the end, instead of the words ** but straight lines
which meet neither way," ought to be read, ** but straight lines in
the same plane which produced meet neither way ;" because, though
in citing this definition in prop. 27, book 1, it was not necessary to
mention the words, **• in the same plane," all the straight lines in the
books preceding this being in the same plane, yet here it was quite
necessary.

PROP. XX. B. XL

In this, near the beginning, are the words, "But if not let BAC
be the greater;" but the angle BAC may happen to be equal to one
of the other two : wherefore this place should be read thus, ** But if
not, let the angle BAC be not less than either of the other two, but
greater than DAB."

At the end of this proposition it is said, " in the same manner it
may be demonstrated," though there is no need of any demonstra-
tion ; because the angle BAC being not less than either of the other
two, it is evident that BAC together with one of them is greater
than the other.

PROP. xxn. B. XI.

And likewise in this, near tBe beginning, it is sard, **^ but if not,
let the angles at B, E, H be unequal, and let the angle at B be
greater than either of those at EH:" which words manifestly show
this iriace to be vitiated, because the angle at B may be eqpal to one
of the other two. They ought therefore to be read thus, " But if
not, let the angles at B, E, H be unequal, and let the angle at B be
not less than either of the other two at E, H: therefore the straight
line AC is not less than either of the two DF, GK."

PROP. xxm. B. XL

The demonstration of this is made something shorter, by not
repeatmg in the third case the things which were demonstrated

t80 NOTES. BOOK XI.

In the first ; and by making use of the constnictfdn which Cam-
panus has given ; but he does not demonstrate the second and third
cases: the construction and demonstration of the third case are
made a little more simple than in the Greek tesrt.

PROP. XXIV. B. XI.

The word ** similar" is added to the enunchition of this proposi-
tion, because the planes containing the solids which are to be de-
monstrated to be equal to one another, in the 25th proposition, ought
to be similar and equal, that the equality of the solids may be ioferred
from prop. C, of this book ; and, in the Oxford edition of Comman-
dine's translation, a corollary is added to prop. 24, to show that the
parallelograms mentioned in this proposition are similar, that the
equality of the solids in prop. 25, may be deduced Orota the 10th def
of book 1 i.

PROP. XXV. and XXVI. B. XI.

In the 25th prop, solid figures which are contained by the same
number of similar and equal plane figures, are supposed to be equal
to one another. And it seems that Theon, or some other editor,
that he might save himself the trouble of demonstrating the solid
figures mentioned in this proposition to be equal to one ianother, has
inserted the 10th def. of this book, to serve instead of a demonstra-
tion ; which was very ignorantly done.

Likewise in the 26th prop, two solid angles are supposed to be
equal : if each of them be contained by three plane angles which
are equal to one another, each to each. And it is strange enough,
that none of the commentators on Euclid have, as far as I know,
perceived that something is wanting in the demonstrations of these
two propositions. Clavius, indeed, in a note upon the 11 th def of
this book, affirms, that it is evident that those solid angles are equal
which are contained by the same number of plane angles, equal to
one another, each to each, because they will coincide, if they be con-
ceived to be placed within one another ; but this is said without any
proof, nor is it always true, except when the solid angles are con-
tained by three plane angles only, which are equal to one another,
each to each : and in this case the proposition is the same with this,
that two spherical triangles that are equilateral to one another, are
also equiangular to one another, and can coincide; which ought not
to be grafted without a demonstration. Euclid does not assume
this in the case of rectilineal triangles, but demonstrates, in prop. 8,
book 1, that triangles which are equilateral to one another are also
equiangular to one another ; and from this their total equality appears
by prop. 4, book 1. And Menelaus, in the fourth prop, of bis Ist
book of spherics, explicitly demonstrates, that spherical triangles
which are mutually equilateral, are also equiangular to one another;
from which it is easy to show that they must coincide, providing
they have their sides disposed in the same order and situation. *

To supply these defects, it was necessary to add the three pro-

BOOK XI. NOTES. 281

positions marked A, B, C to this book. For the 25th, 26th and 28th
propositions of it, and consequently eight pthers, viz. the 27th, 31st,
32d, 33d, 34th, 36th, 37th, and 40th of the same, which depend upon
them, have hitherto stood upon an infirm foundation ; as also the
8th, 12th, cor. of the 17th and 18th of 12th book, which depend upon
the 9th definition. For it has been shown in the notes on def. 10th
of this book, that solid figures ' which are contained by the same
number of similar and equal plane figures, as also solid angles that
are contained by the same number of equal plane angles, are not al-
ways equal to one another.

It is to be observed that Tacquet, in his Euclid, defines equal solid
angles to be such, " as being put within one another do coincide ;"
but this is an axiom, not a definition ; for it is true of all magnitudes
whatever. He made this useless definition, that by it he might de-
monstrate the 36th prop, of this book, without the help of the 35th
of the same ; concerning which demonstration, see the note upon
prop. 36.

PROP. XXVIII. B. XI.

In this it ought to have been demonstrated, not assumed, that the
diagonals are in one plane. Clavius has supplied this defect.

PROP. XXIX. R XI.

There are three cases of this proposition : the first is, when the
two parallelograms opposite to the base AB have a side common to
both ; the second is, when these parallelograms are separated from
one another ; and the third, when there is a part of them common to
both ; and to this last only, the demonstration that has hitherto been
in the Elements does agree. The first case is immediately deduced
from the preceding 28th prop, which seems for this purpose to have
been premised to this 29th, for it is necessary to none but to it, and
to the 40th of this book, as we now have it, to which last it would,
without doubt, have been premised, if Euclid had not made use of it
in the 29th ; but some unskilful editor has taken it away from the
Elements, and has mutilated Euclid's demonstration of the other two
cases, which is now restored, and serves for both at once.

PROP. XXX. B. XI.

In the demonstration of this, the opposite planes of the sc^id OP,
in the figure of this edition, that is of the solid CO in Commandine*s
figure, are not proved to be parallel ; which it is proper to do for the
sake of learners.

PROP. XXXI. B. XL

There are two cases of this proposition : the first is, when the
insisting straight lines are right angles to the bases; the other,
when they are not : the first case divided again into two others, one
of which is, when the bases are equiangular parallelograms ; the
other, when they are not equiangular ; the Greek editor makes no
mention of the first of these two last cases, but has inserted the de-
monstration of it as a part of that of the other ; and therefore should

i83 NOTES. BOOK XI.

have taken notice of it in a corollary ; but we thought it better to
give these two cases separately ; the demonstration also is made
something shorter by following the way Euclid has made use of in
prop. 14, book 6. Besides, in the demonstration of the case in
which the insisting straight lines are not at right angles to the bases,
the editor does not prove that the soUds described in the constroc*
tion are parallelopipeds, which it is not to be thought that Euclid ne-
glected : also the words <* of which the insisting straight lines are
not in the same straight lines," have been added by some unskilful
hand ; for they may be in the same straight lines.

PROP. XXXII. B. XI.

The editor has forgot to order the parallelogram FH to be applied
in the angle FGH equal to the angle LOG, which is necessary.
Clavius has supplied this.

Also, in the construction, it is required to complete the solid of
which the base is FH, and altitude the same with that of the solid
CD : but this does not determine the solid to be completed, since
there may be innumerable solids upon the same base, and of the
same altitude : it ought therefore to be said, ** complete the solid of
which the base is FH, and one of its insisting straight lines is FD ;•*
the same correction must be made in the following proposition, 33.

PROP. D. B. XI.

It is very probable that Euclid gave this proposition a place in the
Elements, since he gave the like proposition concerning equiangular
parallelograms in the 23d, B. 6.

PROP. XXXIV. B. XL

In this the words uv ai 6(ps<fT0)(fM sx eitftv cri rcjv avruv €v6Btosvy *' <]€
which the insisting straight lines are not in the same straight lines,**
are thrice repeated ; but these words ought either to be left out, as
they are by Clavius, or, in place of them, ought to be put, " whettier
the insisting straight lines be, or be not, in the same straight lines :"
for the other case is without any reason excluded ; also the w6rds
6JV ra v^y " of which the altitudes," are twice put for wv oi' 8(pf tf^wtfoi,
" of which the insisting straight lines ;" which is a plain mistake :
for the altitude is always at right angles to the base.

PROP. XXXV. B. XL

The angles ABH, DEM are demonstrated to be right angles in a
shorter way than in the Greek ; and in the same way ACH, DFM
may be demonstrated to be right angles : also the repetition of the
same demonstration, which begins with " in the same manner," is
left out, as it was probably added to the text by some editor ; for
the words, " in like manner we may demonstrate," are not inserted
except when the demonstration is not given, or when it is something
different from the other, if it be given, as in prop. 26, of this book.— -
Campanus has not this repetition.

We have given another demonstration of the corollary, besides

BQpK XI. NOTE*. ^83

the one in the original, by help of which the following 3Wh prop,
may be demonstrated without the 35th.

PROP. XXXVI. B, XI.

Tacquet in his Eudid demonstrates this proposition without the
help of the 35th ; but it is plain, that the solids mentioned in the
Greek text in the enunciation of the proposition as equiangular, are
such that their solid angles are contained, by three plane angles
equal to one another, each to each ; as is evident from the construc-
tion. Now Tacquet does not demonstrate, but assumes these solid
angles to be equal to one another ; for he supposes the solids to be
already made, and does not give the construction by which they are
made : but, by the second demonstration of the preceding corollary,
his demonstration is rendered legitimate likewise in the case where
the solids are constructed as in the text.

PROP. XXXVII. B. XL

In this it is assumed, that the ratios which are triplicate of those
ratios which are the same with one another, are likewise the same
with one another; and that those ratios are the same with one
another, of which the triplicate ratios are the same with one another;
but this ought not to be granted without a demonstration ; nor did
Euclid assume the first and easiest of these two propositions, but
demonstrated it in the case of duplicate ratios, in the 22d prop,
book 6. On this account, another demonstration is given of this
proposition like to that which Euclid gives in prop. 22, book 6, as
Clavius has done.

PROP. XXXVIII. B. XI.

When it is required to dr^w a perpendicular from a point in one
plane, which is at right angles to another plane, unto this last plane,
it is done by drawing a perpendicular from the point to the common
section of the planes ; for this perpendicular will be perpendicular
to the plane by def. 4, of this book: and it would be foolish in this
case to do it by the 11th prop, of the same: but Euclid (17, 12, in
otlier editions), Apollonius, and other geometers, when they have
occasion for this problem, direct a perpendicular to be drawn from
the point to the plane, and conclude that it will fell upon the common
section of the planes, because this is the very same thing as if they
had made use of the construction above mentioned, and then conclu-
ded that the straight line must be perpendicular to the plane; but
is expressed in fewer words. Some editor, not perceiving this,
thought it was necessary to add this proposition, which can never
be of any use to the 1 1th book, and its being near to the end among
propostions with which it has no connexion, is a mark of its having
been added to the text.

PROP. XXXIX. B. XI.

In this it is supposed, that the straight lines which bisect the
sides of the opposite planes, are in one plane, which ought to have
been demonstrated; as is now done.

284 NOTES. BOOK xn.

BOOK XII.

The learned Mr. Moore, professor of Greek in the University
of Glasgow, observed to me, that it plainly appears from Archiraedes's
epistle to Dositheus, prefixed to his books of the Sphere and Cylinder,
which epistle he has restored from ancient manuscripts, that Eudoxus
was the author of the chief propositions in this 12th book*

PROP. II. B. XII.

At the beginning of this it is said, " if it be not so, the square of
BD shall be to the square of FH, as the circle ABCD is to some
space either less than the circle EFGH, or greater than it." And the
like is to be found near to the end of this proposition, as also in
prop. 5, 11, 12, 18, of this book: concerning which, it is to be ob-
served, that, in the demonstration of theorems, it is sufl5cient, in this
and the like cases, that a thing made use of in the reasoning can
possibly exist, providing this be evident, though it cannot be exhibit-
ed or found by a geometrical construction ; so, in this place it is
assumed, that there may be a fourth proportional to these three
magnitudes, viz. the squares of BD, FH, and the circle ABCD;
because it is evident that there is some square equal to the circle
ABCD though it cannot be found geometrically : and to the three
rectilineal figures, viz. the squares of BD, FH, and the square which
is equal to the circle ABCD, there is a fourth square proportional ;
because to the three straight lines which are their sides, there is a
fourth straight line proportional, (12. 6.) and this fourth square, or a
space equal to it, is the space which in this propositon is denoted by
the letter S : and the like is to be understood in the other places
above cited; and it is probable that this .has been shown by Euclid,
but left out by some editor ; for the lemma, which some unskilful
hand has added to this proposition, explains nothing of it.

PROP. III. B. XII.

In the Greek text and the translations, it is said, <* and because
the two straight lines BA, AC which meet one another," &c. ; here
the angles BAC, KHL are demonstrated to be equal to one another
by 10th prop. B. II. which had been done before : because the trian-
gle EAG was proved to be similar to the triangle KHL : this repeti-
tion is left out, and the triangles BAC, KHL are proved to be similar
in a shorter way by prop. 21, B. 6.

PROP. IV. B. XII.
A few things in this are more fully explained than in the Greek text.

PROP. V. B. XII.

In this, near to the end, are the words ws gfjwrjofe^sv eSeix^ " as was
before shown ;" and the same are found again in the end of prop. 18,
of this book : but the demonstration referred to, except it be the
useless lemma annexed to the 2d prop., is no where in these Ele-
ments, and has been perhaps left out by some editor who has forgot
to cancel those words also.

BOOK Xfl.

MOTES.

265

PROP. VI. B.XIL

A shorter demonstration is given of this; and that which is in the
Greek text may be made shorter by a step than it is, for the author
of it makes use of the 22d prop, of B. 5, twice; whereas once would
have served his purpose : because that proposition extends to any
number of magnitudes which are proportionals taken two and two,
as well as to three which are proportional to other three.

COR. PROP. vm. B. XII.

The demonstration of this is imperfect, because it is not shown
that the triangular pyramids into which those upon multangular
bases are divided, are similar to one another, as ought necessarily
to have been done, and is done in the like case in prop. 12th of
this book. The full demonstration of the corollary is as follows :

Upon the polygonal bases ABODE, FGHKL, let there be simi-
lar and similarly situated pyramids which have the points M, N
for their vertices: the pyramid ABCDEM has to the pyramid
FGHELN the triplicate ratio of that which the side AB has to
the homologous side FG.

Let the polygons be divided into the triangles ABE, EBC, ECD ;
FGL, LGH, LHK, which are similar (20. 6.) each to each, and
because the pyramids are similar, therefore (11. def. 11.) the tri-
angle ELAM is simUar to the triangle LFN, and the triangle ABM to
FGN : wherefore (4. 6.) ME is to EA, as NL to LF ; and as AE to
EB, so is FL to LG, because the triangles EAB, LFG are similar;
therefore, ex sequcUi, as ME to EB, so is NL to LG : in like man-
ner it may be shown that EB is to BM, as LG to GN ; therefore
again, ex ssquali, as EM to MB, so is LN to NG ; wherefore the tri-

angles EMB, LNG having their sides proportionals, are (5. 6.) equi-
angular and similar to one another : therefore the pyramids which
have the triangles EAB, LEG for their bases, and the points M,
N for their vertices, are similar (11. def 11.) to one another, for
their solid angles are (b. 11.) equal, and the solids themselves are
contained by the same number of similar planes: in the same man-
ner, the pyramid EBCM may be shown to be similar to the pyra-
mid LGHN, and the pyramid ECDM to LHKN. And because
the pyramids EABM, LFGN are similar, and have triangular

^iM NOTSa. BOOK x»«

bases, the pyramid EABM has (8. 12.) to LFGN the triplicate
ratio of that which EB has to the homologous side LG. And, in the
same manner, the pyramid EBCM has to the pyaamid LGHN the
triplicate ratio of that which EB has t9 LG. Therefore as the pyra-
mid EABM is to the pyramid LFGN, so is tlie pyramid EBCM to the
pyramid LGHN. In the like manner, as the pyramid EBCM is to
LGHN, so is the pyramid ECDM to the pyramid LHKM. And as
one of the antecedents is to one of the consequents, so are aU the
antecedents to all the consequents : therefore as the pyramid EABM
to the pyramid LFGN, so is the whole pyramid ABCDEM to the
whole pyramid FGHKLN : and the pyramid EABM has to the pyra-
mid LFGN the triplicate ratio of that which AB has to FG ; therefore
the whole pyramid has to the whole pyramid the triplicate ratio of
that which AB has to the homologous side FG. Q. E. D.

PROP. XL and XH. B. XIL

The order of the letters of the alphabet is not observed in these
two propositions according to Euclid's manner, and is now re-
storedh; by which means, the first part of prop. 18 may be demon-
strated in the same words with the first part of prop. 11 : on this
account the demonstration of that first part is left out, and assumed
from prop. 1 1.

PROP. XIL B. XIL

In this proposition, the common section of a plane parallel to
the bases of a cylinder, with the cylinder itself, is supposed to be a
circle, and it was thought proper briefly to demonstrate it; from
whence it is sufficiently manifest, that this plane divides the cy-
linder into two others ; and the same thing is understood to be
supplied in prop. 14.

PROP. XV. B. XIL

"And complete the cylinders AX, EO," both the enunciation and ex-
position of the proposition represent the cylinders as well as the
cones, as already described : wherefore the reading ought rather to
be, " and let the cones be ALC, ENG ; and the cylinders AX, EO."

The first case in the second part of the demonstration is wanting;
and something also in the second case of that part, before the re[)eti-
tion of the construction is mentioned ; which are now added.

PROP. xvn. B. xn.

In the enunciation of this proposition, the Greek words sig rnv
fiiCi^oya (fqxxfgav (frs^sov iroKvsSgov &^J^a4'CCi |X«) -s^auoy rr^g eXafl'o'ovo^ (f(poiigac
KBf/ra Ti)v $^i(pavsiay are thus translated by Commandine and others,
"in majori solidum polyhedron describere quod minoris sphsrae
superficiem non tangat;" that is, "to describe in the greater sphere
a solid polyhedron which shall not meet the superfiicies of the
lesser sphere ;" whereby they refer the words xara Tr^v g-ri(pavciav to
these next to them Tv^g sKcKftfovog (f^tgetg. But they ought by no
means to be thus translated; for the solid polyhedron doth not
only meet the superficies of the leaser sphere, but pervades the

BOOK XII. NOTB0. S87

whole of that sphere; therefore the aforesaid words are to be re-
ferred to <ro (Trs^sov qroXus^^ev, and ought thus to be translated, vis«
to descibe in the greater sphere a solid polyhedron whose super-
ficies shall hot meet the lesser sphere ; as the meaning of the pro-
position necessarily requires.

The demonstration of the proposition is spoiled and mutilated ;
for Some easy things are very explicitly demonstrated, while others
not so obvious are not sufficiently explained: for example, when
it is affirmed^ that the square of KB is greater than the double
of the square of BZ, in the first demonstration, and that the angle
BZK is obtuse, in the second ; both which ought to have been de*
monstrated. Besides, in the first demonstration it is said, *' draw
Kn from the point K perpendicular to BD ;" whereas it ought to
have been said ** join KV," and it should have been demonstrated
that KV is perpendicular to BD : for it is evident from the figure
in Hervagius's and Gregory's editions, and ' from the words of the
demonstration, that the Greek editor did not perceive that the
perpendicular drawn from the point K to the straight line BD
must necessarily fall upon the point V, for in the figure it is made
to fall upon the point Q, a different point from V, which is like-
wise supposed in the demonstration. Commandine seems to have
been aware of this : for in this figure he makes one and the same*
point with the two letters V, Q. ; and before Commandine, the
learned John Dee, in the commentary he annexes to this proposi-
tion in Henry Billinsley's translation of the Elements, printed at
London, ann. 1570, expressly takes notice of this error, and gives
a demonstration suited to the construction in the Greek text, by
which he shows that the perpendicular drawn from the point K
to BD, must necessarily fall upon the point V.

Likewise it is not demonstrated, that the quadrilateral figures
SOPT, TPRY, and the triangle YRX, do not meet the lesser
sphere, as was necessary to have been done : only Clavius, as far
as I know, has observed this, and demonstrated it by a lemma,
which is now premised to this proposition, something altered and
more briefly demonstrated.

In the corollary of this proposition, it is supposed that a solid
polyhedron is described in the other sphere similar to that which
is described in the sphere BCDE ; but, as the construction by
which this may be done is not given, it was thought proper to
give it, and to demonstrate, that the pyramids in it are similar to
those of the same order in the solid polyhedron described in the
sphere RCDE.

From the preceding notes, it is sufficiently evident how much
the elements of Euclid, who was a most accurate geometer, have
been vitiated and mutilated by ignorant editors. The opinion
which the greatest part of learned men have entertained concern-
ing the present Greek edition, viz. that it is very little or nothing
dififbrent firom the genuine work of Euclid, has without donbt de-
ceived them, and made them less attentive and accurate in exa-
mining that edition ; whereby several errors, some of them gross
enough, have escaped their notice, from the age in which Theon

288 NOTES. BOOK XII.

lived to this time. Upon which account there is some ground to
hope that the pains we have taken in correcting those errors, and
freeing the Elements as far as we could from blemishes, will not
be unacceptable to good judges, who can discern when demon-
strations are legitimate, and when they are not.

The objections which, since the first edition, have been made
against some things in the notes, especially against the doctrine
of proportionals, have either been fully answered in Dr. Barrow's
Lect. Mathemat. and in these notes; or are such, except one
which has been taken notice of in the note on prop. 1, book 11, as
show that the person who made them has not sufficiently consi-
dered the things against which they are brought ; so that it is not
necessaiy to make any further answer to these objections and
others like them against Euclid's definition of proportionals: of
which definition Dr. Barrow justly says, in page 297 of the above
named book, that **Nisi machinis impulso validioribus aetemum
' persistet inconcussa.*'

P1NI3.*

EUCLID'S DATA.

IN THIS EDITION

SEVERAL ERRORS ARE CORRECTED,

AND

SOME PROPOSITIONS ADDED.

, BY ROBERT SIMSON, M. D.

, EMEEITUf FEOmSOR Ol MATHEMATICS IN THB UNIVXEHTY OF OLAMOW.

PREFACE.

EUCLID'S DATA is the first in order of the books written by the ancient
geometers to fiicilitate and promote the method of resolution or analysis. In
Uie fi^eneral, a thing is said to be given which is either actually exhibited, or
can be* found out, that is, which is either known by hypothesis, or that can
be demonstrated to be known; and the propositions in the book of Euclid's
Data show what thinffs can be found out or known from those that by hy-
pothesis are already known ; so that in the analysis or investigation of a
problem, from the things that are laid down to be known or given, by the
help of these propositions other things are demonstrated to be given, and
from these, other things are again shown to be given, and so on, until that
which was proposed to be fi)um out in the problem is demonstrated to be
given, and when this is done, the problem is solved, and its composition is
made and derived from the compositions of the Data which were made use
of in the analysis. And thus the Data of Euclid are of the most general and
necessary use in the solution of problems of every kind.

Euclid is reckoned to be the author of tiie Book of the Data, both by the
ancient and modem geometers ; and there seems to be no doubt of his having
written a book on this subject, but which in the course of so many ages, has
been much vitiated by unskilful editors in several places, both in tlie order
of the propositions, and in the definitions and demonstrations themselve&
To correct the errors which are now found in it, and bring it nearer to the
accuracy with which it was, no doubt, at first written by Euclid, is the de-
sign of this edition, that so it may be rendered more usefiil to geometers, at
least to beginners who desire to learn the investigatory method of the anci-
ents. And for their sakes, the compositions of most of the Data are sub-
joined to their demonstrations, that the compositions of problems solved by
help of the Data may be the more easily made.

Marinus the philosopher's pre&ce, which, in the Greek editions, is pre-
fixed to the Data, is here left out, as being of no use to understand them.
At the end of it, he says, that Euclid has not used the synthetical, but the
analjrtlcal method in delivering them ; in which he is quite mistaken ; for,
in the analysis of a theorem, the thing to be demonstrated is assumed in the
analysis; but in the demonstrations of the Data, the thing to be demonstra-
ted, which is, that something or other is given, is never once assumed in
the demonstration, ilrom which it is manifest, that every one of them is
demonstrated synthetically ; tliough, indeed, if a proposition of the Data be
turned into a problem, for example the 84th or 85th m the former editions,
which here are |he 85th and 86th, the demonstration of the proposition be-
comes the analysis of the problem.

Wherein this edition differs from the Greek, and the reaFons of the alter-
ations from it, will be shown in the notes at the end of* the Data.

EUCLID'S DATA,

DEFINITIONS.

Spaces, lines, and angles, are said to be given in magnitude, when
equals to them can be found.

A ratio is said to be given, when a ratio of a given magnitude to
' a given magnitude which is the same ratio with it can be found.

Rectilineal figures are said to be given in species, which have
each of their angles given, and the ratios of their sides given.

IV.
Points; lines, and spaces, are said to be given in position, which
have always the same situation, and which are either actually
exhibited, or can be found.

An angle is said to be given in position, which is contained by
straight lines given in position.

A circle is said to be given in magnitude, when a straight line
from its centre to the circumference is given in magnitude.

VI.
A circle is said to be given in position and magnitude, the centre
of which is given in position, and a straight line from it to the
circumference is given in magnitude.

VII.
Segments of circles are said to be given in magnitude, when the
angles in them, and their bases, are given in magnitude.

Vffl.
Segments of circles are said to be given in position and magni-
tude, when the angles in them are given in magnitude, and their
bases are given both in position and magnitude.

IX.
A magnitude is said to be greater than another by a given mag-
nitude, when this given magnitude being taken from it, the re-
mainder is equal to the other magnitude.

A magnitude is said to be less than another by a given magni-
tude, when this given magnitude being added to it, the whole
is equal to the other magnitude.

294

EUCLID S DATA.

PROPOSITION I.

The ratios of given magnitudes to one another are given-f

Let A, B be two given.magnitudes, the ratio of A to B is given.

Because A is a given magnitude, there may
(1. def. Dat.) be found one equal to it ; ]et this
be C, and because B is given, one equal to it
may be found ; let it be D; and since A is equal
to C, and B to D; therefore (7. 5.) A is to B,
as C to D ; and consequently the ratio of A to
B is given, because the ratio of the given mag-
nitudes C, D, which is the same with it, has A fe 6 D
been found.

PROP. II.

Jp a given magnitude has a given ratio to another magnitude,
•^ and if unto the two magnitudes, by which the given ratio is
exhibited, and the given magnitude, a fourth proportional can be
found;" the other magnitude is given.f

Let the given magnitude A have a given ratio to the magni-
tude B ; if a fourth proportional can be found to the three xoagnl*
tudes above named, B is given in magnitude.

Because A is given, a magnitude may be
found equal to it (1. def.); let this be C: and
because the ratio of A to B is given, a ratio
which is the same with it may be found; let
this be the ratio of the given magnitude E to
the given magnitude F: unto the magnitudes
E, F, C find a fourth proportional D, which, by
the hypothesis, can be done. Wherefore be-
cause A is to B, as E to F ; and as E to F, so
is C to D; A is (11. 5.) to B, as C to D. But
A is equal to C; therefore (14. 5.) B is equal to D. The magni-
tude B is therefore given (1. def.) because a magnitude D equal
to it has been found.

The limitation within the inverted commas 19 not in the Greek
text, but is now necessarily added ; and the same must be under-
stood in all the propositions of the book which depend upon this
, second proposition, where it is not expressly mentioned. See the
note upon it.

BCD
E F

• The figure! in the mar^ia show the number of the propositions in the other
editions,
t See Note.

£tlCLID*S DAtA. 296

PROP. ni.

If any given magnitudes be added together, their sum shall be
given.

Let any given magnitudes AB, BC be added together, their sum
AC is given.

Because AB is given, a magnitude equal to it may be found
(i. def.); let this be DE: and because
BC is given, one equal to it may be A B C

found; let this be EF: wherefore, be- (

cause AB is equal to DE, and BC equal

to EF; the whole AC is equal to the D E F

whole DF: AC is therefore given, be- 1 '.

cause DF has been found, which is equal to it.

PROP. IV.

If a given magnitude be taken from a given magnitude, the
remaining magnitude shall be given.

From the given magnitude AB, let the given magnitude AC
be taken; the remaining magnitude CB is given.

Because AB is given, a magnitude equal to it may (1. def.) be
found; let this be DE: and because
AC is given, one equal to it may be A C B

found; let this be DF; wherefore 1 — —

because AB is equal to DE, and AC

to DP; the remainder CB is equal D F E

to the remainder FE. CB is therefore

given (I. def), because FE, which is equal to it, has been found.

PROP. V. la.

Ir of three magnitudes, the first together with the second be
given, and also the second together with the third ; either the
first is equal to the third, or one of them is greater than the
Oth^ by a given magnitude.*

Let AB, Be, CD be three magnitudes, of which AB together
with BC, that is AC, is given ; and also BC together with CD,
that is, BD, is given. Either AB is equal to CD, or one of them
is greater than the other by a given magnitude.

Because AC, BD are each of them given^ they are either
equal to one another, or not eqfLa.1 First,
let them be equal, and because AC is A B CD

equal to BD, take away the common part — j— — — |— —
BC; therefore the remainder AB is equal
to the remainder CD.

But if they be unequal, let AC be greater than BD, and make
CE equal to BD. Therefore CE is given, because BD is given.
And the whole AC is given; there-

*>fe (4. dat) AE the remainder is A E B CD
given. And because EC is equal to [ 1

♦See Note.

296 Euclid's data,

BD, by taking BO from both, the remainder EB is equal to the
remainder CD. And AE is given; wherefore AB exceeds £IB»
that is CD, by the given magnitude AE.

PROP. VI.

If a magnitude has a given ratio to a part of it, it shall also
have a given raito to the remaining part of it*

Let the magnitude AB have a given ratio to AC a part of it;
it has also a given ratio to the remainder BC.

Because the ratio of AB to AC is given, a ratio may be found
(2. def.) which is the same to it : let this be the ratio of DEI a
given magnitude to the given magnitude
DF. And because DE, DF are given, A C B

the remainder FE is (4. dat.) given : and — |

because AB is to AC, as DE to DF, by D F E

conversion (E. 5.) AB is to BC, as DE

to EF. Therefore the ratio of AB to BC is given, because the
ratio of the given magnitudes DE, EF, which is the same with it,
has been found.

Cor. From this it follows, that the parts AC, CB have a given
ratio to one another: because as AB to BC, so is DE to £F;
by division (17. 5.) AC is to CB, as DF to FE : and DF, PE are
given ; therefore (2. def.) the ratio of AC to CB is given.

PROP. vn. 6.

If two magnitudes which have a given ratio to one another,
be added together: the whole magnitude shall have to each of
them a given ratio.*

Let the magnitudes AB, BC which have a given ratio to one
another, be added together; the whole AC has to each of the
magnitudes AB, BC a given ratio. .

Because the ratio of AB to BC is given, a ratio may be found
(2. def.) which is the same with it; let this be the ratio of the
given magnitudes DE, EF: and because

DE, EF are given, the whole DF is given A B C

(3. dat.) : and because as AB to BC, so is -| ■

DE to EF ; by composition (18. 6.), AC is
to CB, as DF to FE ; and by conversion (E. D E F

5.), AC is to AB, as DF to DE; wherefore 1

because AC is to each of the magnitudes AB, BC, as DF to each
of the others DE, EF ; the ratio of AC to each of the magnitudes
AB, BC is given (2. def).

PROP. vm. 7.

If the given magnitude be divided into two parts which have
a given ratio to one another, and if a fourth proportional can be

* See Note.

EUCUD'S DATA.

297

found to the sum of the two magnitudes by which the given ratio
is exhibited, one of them, and the given magnitude; each of the
parts is given.*

Let the giv^i magnitude AB be divided into the parts AC, CB
which have a given ratio to one another ; if a fourth proportional
can be found to the above named magni-
tudes ; AC and CB are each pf them given. A C B

Because the ratio of AC to CB is given, —
the ratio of AB to BC is given (7. dat.) ;
therefore a ratio which is the same with it D P E

can be found (2. def.); let this be the ratio 1

of the given magnitudes, DE, EP : and be*

cause the given magnitude AB has to BC A C B

the given ratio of DE to EP, if unto DE, (

EF, AB a fourth proportional can be

found, this which is BC is given (2. dat.) ; D P E

and because AB is given, the other part 1

AC is given (4. dat.)

In the same manner, and with the like limitation, if the differ-
ence AC of two magnitudes AB, BC which have a given ratio be
given ; each of the magnitudes AB, BC is given.

PROP. EX. 8.

Magnitudes which have given ratios to the same magnitude^
have also a given ratio to one another.

Let A, C have each of them a given ratio to B; A has a given
ratio to C.

Because the ratio of A to B is given, a ratio which is the same
to it may be found (2. def) ; let this be the ratio of the given mag-
nitudes D, E : aod because the ratio of B to C is given, a ratio
which is the same with it may be found (2. def.) ; let this be the
ratio of the given magnitudes P, G :
to F, G, E find a fourth proportional
H, if it can be done; and because
as A is to B, so is D to E ; and as B
to C, so is (P to G, and so is) E to
H ; ex ssqualh as A to C, so is D to
H: therefore the ratio of A to C is A B
given (2. def) because the ratio of
the given magnitudes D and H,
wliich is the same with it, has been
found: but if a fourth proportional
to F, G, E cannot be found, then it
can only be said that the ratio of A to C is compounded of the
ratios of A to B, and B to C, that is, of the given ratios of D to £,
and F to G.

E H
G

* See Note.
38

298 EUCLlD^a DATA.

PROP X. 9.

If two or more magnitudes have given ratios to one another,
and if they have given ratios, though they be not the same, to
some other magnitudes ; these other magnitudes shall also have
given ratios to one another.

Let two or more magnitudes A, B, C have given ratios to one
another ; and let them have given ratios, thou^ they be not the
same, to some other magnitudes D, £, F; the magnitudes D, £,
P have given ratios to one another.

Because the ratio of A to B is given, and likewise the iratio of
A to D ; therefore the ratio of D to

B is given (9. dat.): but the ratio of A———- D

B to E is given, therefore (9. dat.) B— — E

the ratio of D to E is given : and be- C F— :

cause the ratio of B to C is given, ^

and also the ratio of B to E ; the ratio of E to C is given (9. dat.) :
and the ratio of C to F is given ; 'therefore the ratio of E to F is
given ; D, E, F have therefore given ratios to one another.

PROP. XL 23.

If two magnitudes have each of them a given ratio to another
magnitude, both of them together shall have a given ratio to that
other.

Let the magnitudes AB, BC have a given ratio to the magni-
tude D ; AC has a given ratio to the same D.

Because AB, BC have each of them p

a given ratio to D, the ratio of AB to . . ^

BC is given (9. dat.) : and by composi- '

tion the ratio of AC to CB is given; j^ ^^^

(7. dat.) : but the ratio of BC to D is

given ; therefore (9. dat.) the ratio of AC to D is given.

PROP. xn. 23.

If the whole have to the whole a given ratio, and the parts
have to the parts given, but not the same, ratios, every one of
them, whole or part, shall have to every one a given ratio.*

Let the whole AB have a given ratio to the whole CD, and the
parts AE, EB have given, but not the same, ratios to the ports
CF, FD, every one shall have to every one, whole or part, a given
ratio.

Because the ratio of AE to CP is given, as AE to CF, so
make AB to CG; the ratio therefore of AB to CG is given;
wherefore the ratio of the remainder EB to the remainder FG is

* See Note.

EUCLID S DATA.

2«&

G D

givim, because it is the same (19. 5.) with the ratio of AB to OG:
and the ratio of EB to PD is given, where- a r» -o

fore the ratio of FD to FG is given (9. _
dat); and by conversion, the ratio of Fp
to DG is given (6. dat.) : and because AB ^

has to each of the magnitudes CD, CG a

given ratio, the ratio of GD to CG is
given (9. dat) ; and therefore (6. dat.) the ratio of CD to DG is
given : but the ratio of GD to DF is given, wherefore (9. dat) the
ratio of CD to DF is given, and consequent]};^ (cor. 6. dat.) the
ratio of CF to FD is given ; but the ratio of CF to AE is given,
as also the ratio of FD to EB, wherefore (10. dat) the ratio of AE
to EB is given ; as also the ratio of AB to each of them (7. dat) :
the ratio therefore of one to every one is given.

PROP. xm.

24.

If the first of three proportional straight lines has a given
ratio to the third, the first shall also have a given ratio to the
second.* »

Let A, B, C be three proportional straight lines, that is, as A
to B, so is B to C ; if A has to C a given ratio, A shall also have
to B a given ratio.

Because the ratio of A to C is given, a ratio which is the same
with it may be found (2. def.) ; let this be the ratio of the given
straight lines D, E;%nd between D and E find a (13.6.) mean
proportional F ; therefore the rectangle contained by D and E is
equal to the square of F, and the rectangle D, E
is given, because its sides D, E are given ; where-
fore the square of F, and the straight line F is
given: and because as A is to C, so is D to E;
but as A to C, so is (2. cor. 20. 6.) the square of
A to the square of B; and as D to E, so is (2. cor.
20. 6.) the square of D to the square of F : there-
fore the square (11. 5.) of A is to the square of B,
as the square of D to the square of F : as there-
fore (22. 6.) the straight line A to the straight
line B, so is the straight line D to the straight
line F : therefore the ratio of A to B is given (2.
def.), because the ratio of the given straight lines
D, F, which is the same with it, has been found.

B C

D F

E
I

PROP. XIV.

Ir a magnitude together with a given magnitude has a given
ratio to another magnitude ; the excess of this other magnitude
above a given magnitude has a given ratio to the first magnitude:
and if the excess of a magnitude above a given magnitude has
a given ratio to another magnitude; this other magnitude together

* See Note.

300 EUr uO's DATA.

with a given magnitude has a given ratio to the first mag-
nitude.*

Let the magnitude AB together with the given magnitude BE,
that is, AE, have a given ratio to the magnitude CD ; the excess
of CD above a given magnitude has a given ratio to AB.

Because the ratio of AE to CD is given, as AE to CD, so
make BE to FD; therefore the ratio of BE to FD is given, and
BE is given; wherefore FD is given . BE

(2. dat,) : and because as AE to CD, i

so is BE to FD, the remainder AB is "" ' '

(19. 6.) tp the remainder CF, as AE C F D

to CD : but the ratio of AE to CD is "

given, therefore the ratio of AB to CF is given ; that is, CF, the
excess of CD above ^ the given magnitude FD, has a given ratio
to AB.

Next, Let the excess of the magnitude AB above the given
magnitude BE, that is, ]et AE have a given ratio to the magni-
tude CD : CD together with a given magnitude has a given ratio
toAR «

Because the ratio of AE to CD is given, as AE to CD, so
make BE to FD; therefore the ratio ^^ a E B

BE to FD is given, and BE is given, i

wherefore FD is' given (2. dat.). And '

because as AE to CD, so is BE to FD, C D F

AB is to CF, as (12. 5.) AE to CD : but

the ratio of AE to CD is given, therefore the ratio of AB to CP
is given : that is, CF, which is equal to CD together with the
given magnitude DF, has a given ratio to AB.

PROP. XV. R

If a magnitude, together with that to which another magnitude
has a given ratio, be given; the sum of this other, and that to
which the first magnitude has a given ratio, is given.*

Let AB, CD be two magnitudes, of which AB together with
BE, to wMch CD has a given ratio, is given; CD is given, to-
gether with that magnitude to which AB has a given ratio.

Because the ratio of CD to BE is given, as BE to CD, so make
AE to FD ; therefore the ratio of AE to FD is given, and AE is
given, wherefore (2. dat.) FD is given : . BE

and because as BE to CD, so is AE to ^ ,

FD : AB is (cor. 19. 5.) to FC, as BE to '

CD : and the ratio of BE to CD is given, F CD

wherefore the ratio of AB to FC is

given : and FD is given, that is CD together with FC, to which
AB has a given ratio, is given.

« See Note.

Euclid's data. ' 301

PROP. XVI. 10.

If the excess of a magnitude, above a given magnitude, has a
given ratio to another magnitude ; the excess of both together
above a given magnitude shall have to that other a given ratio :
and if the excesk of two magnitudes together above a given
magnitude^ has to one of them a given ratio ; either the excess
of the other above a given magnitude has to that one a given
ratio, or the other is given together with the magnitude to which
that one has a given ratio.*

Let the excess of the magnitude AB above a given idagnitude,
liave a given ratio to the magnitude BC ; the excess of AC, both
of them together, above the given magnitude, has a given ratio
to BC.

Let AD be the given magnitude, the excess of AB above
which, viz. DB, has a given ratio to . D B C

BC: and because DB, has a given ra- , , _^

tio to BC, the ratio of DC toCB is ' '

given (7. dat.), and AD is given; therefore DC, the excess of

AC above the given magnitude AD, has a given ratio to BC.

Next, Let the excess of two magnitudes AB, BC together,
above a given magnitude, have to one . • n p i? r«

of them BC a given ratio ; either the ^ | j

excess of the other of them AB above

the given magnitude shall have to BC a given ratio ; or AB is

given, together with the magnitude to which BC has a given

ratio.

Let AD be the given magnitude, and first let it be less than
AB; and because DC, the excess of AC above AD has a given
ratio to BC, DB has (cor. 6. dat.) a given ratio to BC ; that Is,
DB, the excess of AB above the given magnitude AD, has a
given ratio to BC.

But let the given magnitude be greater than AB, and make AE
equal to it ; and because EC, the excess of AC above AE, has to
BC a given ratio, BC has (6. dat.) a given ratio to BE ; and be-
cause AE is given, AB together with BE, to which BC has a
given ratio, is given.

PROP. XVII. 11.

If the. excess of a magnitude above a given magnitude have
a given ratio to another magnitude ; the excess of the same first
magnitude above a given magnitude, shall have a given ratio to
both the magnitudes together. And if the excess of either of
two magnitudes above a given magnitude have a given ratio to
both magnitudes together ; the excess of the same above a given
magnitude shall have a given ratio to the other.*

«SeeNote.

Euclid's data*

Let the excess of the magnitude AB above a given magnitude
have a given ratio to the magnitude BC : the excess of AB above
a given magnitude has a given ratio to AC.

Let AD be the given magnitude ; and because DB, the excess
of AB above AD, has a given ratio to BC ; the ratio of DC to
DB is given (7. dat): make the ratio of AD to DB the same
with this ratio; therefore the ratio

of AD to DE is given: and AD is A E D B C
given, wherefore (2. dat.) DE, and --, 1 1.

the remainder AE are given: and because as DC to DB, so is
AD to DE, AC is (12. 5.) to EB, as DC to DB ; and the ratio of
DC to DB is given ; wherefore the ratio of AC to EB is given :
and because the ratio of EB, to AC is given, and that AE is
given, therefore EB, the excess of AB above the given magnitude
AE, has a given ratio to AC.

Next, Let the excess of AB above a given magnitude have a given
ratio to AB and BC together, that is, to AC; the excess of AB above
a given magnitude has a given ratio to BC.

Let AE be the given magnitude ; and because EB, the excess
of AB above AE has to AC a given ratio, as AC to EB, so make
AD to DE; therefore the ratio of AD to DE is given, as also
(6, dat.) the ratio of AD to AE: and AE is given wherefore
(2. dat'.) AD is given: aild because, as the whole AC, to the
whole EB, so is AD to DE, the remainder DC is (19. 6.) to the
remainder DB, as AC to EB; and the ratio of AC to EB is
given ; wherefore the ratio of DC to DB is given, as also (cor. 6.
dat.) the ratio of DB to BC: and AD is given; therefore DB, the
excess of AB above a given magnitude AD, has a given ratio to
BO. - .

PROP. XVm. 14.

If to each of two magnitudes, which have a given ratio to
one another, a given magnitude be added ; the wholes shall either
have a given ratio to one another, or the excess of one of them
above a given magnitude shall have a given ratio to the other.

, Let the two magnitudes AB, CD have a given ratio to one an-
other, and to AB let the given magnitude BE be added, and the
given magnitude DF to CD : the wholes AE, CP either have «l given
ratio to one another, or the excess of one of them above a given
magnitude has a given ratio to the other (1. dat.).

Because BE, DF are each of them given, their ratio is given,
and if this ratio be the same with A B E

the ratio of AB to CD, the ratio of

AE to CF, which is the same (12. 5.)

with the given ratio of AB to CD, C D

shall be given. 1-

* See Note.

Euclid's data. 303

But if the ratio of BE to DF be not the same with the ratio of
AB to CD, either it is greater than the ratio of AB to CD, or, by
inversion, the ratio of DF to BE is greater than the ratio of CD
to AB: first, let the ratio of BE to DF A B G E

be greater than the ratio of AB to CD; 1 1

and as AB to CD, so make BG to DF;
therefore the ratio of BG to DF is given; CDF
and DF is given, therefore (2. dat.) BG is | ■

given : and because BE has a greater ratio to DF than (AB to
CD, that is, than) BG to DF, BE is greater (10. 5.) than BG; and
because as AB to CD, so is BG to DF; therefore AG is (12. 6.)
to CF, as AB to CD : but the ratio of AB to CD is given, where-
fore the ratio of AG to CF is given ; and because BE, BG are each
of them given, GE is given : therefore AG, the excess of AE above
a given magnitude GE, has a given ratio to CF. The other case is
demonstrated in the same manner.

PROP. XIX. 16.

If from each of two magnitudes, which have a given ratio to
one another, a given magnitude be taken, the remainders shall
either have a given ratio to one another, or the excess of one of
them above a given magnitude, shall have a given ratio to the
other.

Let the magnitudes AB, CD have a given ratio to one another*
and from AB let the given magnitude AE be taken, and from CD,
the given magnitude CF: the remainders EB, FD shall either
have a given ratio to one another, or the excess of one of them
above a given magnitude shall have a
given ratio to the other. A E B

Because AE, CF are dach of them given,
their ratio is given (1. dat.): and if this
ratio be the same with the ratio of AB to C F D

CD, the ratio of the remainder EB to the 1

remainder FD, which is the same (19. 5.) with the given ratio of AB
to CD, shall be given.

But if the ratio of AB to CD be not the same with the ratio of
AE to CF, either it is greater than the ratio of AE to CF, or, by
inversion, the ratio of CD to AB is greater than the ratio of CF
to ABt First, let the ratio of AB to CD be greater than the ratio
of AE to CF, and as AB to CD, so make AG to CF ; therefore
the ratio of AG to CF is given, and
CP is given, wherefore (2. dat.) AG is A EG B

given : and because the ratio of AB to 1 — j

CD, that is, the ratio of AG to CF, is

lareater than the ratio of AE to CF; C F D

AG is greater (10. 6.) than AE: and

AG, AE are given, therefore the remainder EG is given ; and as
AB to CD, so is AG to CF, and so is (19. 5.) the remainder GB
to the remainder FD ; and the ratio of AB to CD is given : where-
fore the ratio of GB to FD is given ; therefore GB, the excess of

304 Euclid's data.

EB above a given magnitude EG, has a given ratio to FD. In the
same manner the other case is demonstrated.

PROP. XX. 16.

If to one of two magnitudes which have a given ratio to oiie
another, a given magnitude be added, and from the other a given
magnitude be taken ; the excess of the sum above a given mag-
nitude shall have a given ratio to the remainder.

Let the two magnitudes AB, CD have a given ratio t6 one ano-
ther, and to AB, let the given magnitude EA be added, and from
CD let the given magnitude CF be taken ; the excess of the sum
EB above a given magnitude, has a given ratio to the remainder
FD.

Because the ratio of AB to CD is given, make as AB to CD^
so AG to CF: therefore the ratio of AG to OF is given, and CF
is given, wherefore (2. dat) AG is
given; and EA is given, therefore E A G B

the whole EG is given : and because — — -| 1.

as AB to CD, so is AG to CF, and

so is (19. 5.) the remainder GB to C F D

the remainder FD; the ratio of GB ■

to FD is given, and EG is given, therefore GB, the excess of the
sum EB above the given magnitude EG, has a given ratio to the
remainder FD.

PROP. XXL C.

If two magnitudes have a given ratio to one another, if a
given magnitude be added to one of them, and the other be taken
from a given magnitude ; the sum, together with the magnitude
to which the remainder has a given ratio, is ' given ; and the
remainder i^ given together with the magnitude to which the sum
has a given ratio.*

Let the two magnitudes AB, CD have a given ratio to one ano-
ther ; and to AB let the given magnitude BE be added, and let
CD be taken from the given magnitude FD : the sum AE is given,
together with the magnitude to which the remainder FC^as a
given ratio.

Because the ratio of AB to CD is given, make as AB to CD,
so GB to FD : therefore the ratio of GB to FD is given, and FD
is given, wherefore GB is given (2. dat) ;
and BE is given; the whole GE is G A B E

therefore given; and because as AB to 1 [-.■■■

CD, so is GB to FD, and so is (19. 5.)

GA to FC; the ratio of GA to FC is F C D

given: and AE together with GA is — —

* See Note.

E0CL1D*8 DATA. 305

given, because G£ is given ; therefore the sum AE together with
QA, to which the remauider FC has a given ratio, is given. The
second part is manifest from prop. 15.

PROP. xkn. D.

If two nf)agnitudes have a given ratio to one another, if from
one of them a given magnitude be talien, and the other be taken
from a given magnitude ; each of the remainders is given, to-
gether with the magnitude to which the other remainder has
a given ratio.*

Let the two magnitudes AB, CD have a given ratio to one another,
and from AB let the given magnitude AE be tal^en, and let CD be
taken from the given magnitude CF : the remainder EB is given, to-
gether with the magnitude to which the other remainder DF has a
given ratio.

Because the ratio of AB to CD is given, make as AB to CD, so
AG to CF : the ratio of AG to CF is therefore given, ajid CF is given,
wherefore (2. dat.) AG is given ; and . „ „ q

AE is given, and therefore the re- , .

mainder EG is given; and because "' '"

as AB to CD so is AG to CF : and so p . _

is (19, 5.) the remainder BG to the ^ ^ , ^

remainder DF; the ratio of BG to DF " '

is given : and EB together with BG is given, because EG is given :
therefore the remainder EB together with BG, to which DF the other
remainder has a given ratio, is given. The second part is plain from
this and prop. 15.

PROP. XXm. 20.

If from two given magnitudes there be taken magnitudes
which have a given ratio to one another, the remainders shall
either have a given ratio to one another, or the excess of one of
them above a given magnitude shall have a given ratio to the
other.*

Let AB, CD be two given magnitudes, and from them let the
magnitudes AE, CF, which have a given ratio to one another, be
taken ; the remainders EB, FD either have a given ratio to one an-
other ; or the excess of one of them above a given magnitude has a
given ratio to the other.

Because AB, CD are each of them A E B

given, the ratio of AB to CD is given :

and if this ratio be the same with the

ratio of AE to CF, then the remainder C F D

KB has (19. 6.) the same given ratio

to the remamder FD.

* Sec Note.
39

30& BUCLID^S DATA.

But If the ratio of AB to CD be not the same with the ratio of
AE to CF, it is either greater than it, or, by inversion, the ratio
of CD to AB is greater than the ratio of CP to AE : first, let the
ratio of AB to CD be greater ^lan the ratio of AE to CF ; and as
AE to CF, so make AG to CD ; therefore the ratio of AG to CD
is given, because the ratio of AE to CF is given ; and CD is given,
wherefore (2. dat.) AG is given; and because the ratio of AB to
CD is greater than the ratio of (AE to
CF, that is, than the ratio of) AG to A E G B

CD; AB is greater (10. 5.) than AG: — )- ^|

and AB, AG are given; therefore the

remainder BG is given: and because C P D

as AE to CF, so is AG to CD, and so 1 '—

is (19. 5.) EG to FD; the ratio of EG to FE is given : andGBis
given ; therefore EG, the excess of EB above a given magnitude
GB, has a given ratio to FD. The other case is shown in the
same way.

PROP. XXIV. 18.

If there be three magnitudes, the first of which has a given
ratio to the second, and the excess of the second above a given
magnitude has a given ratio to the third ; the excess of the first
above a given magnitude shall also have a given ratio to tlie
third.*

Let AB, CD, E, be the three magnitudes of which AB has a
given ratio to CD ; and the excess of CD above a given magnitude
has a given ratio to E : the excess of AB above a given magnitude
has a given ratio to E.

Let CF be the given magnitude, the excess of CD above which,
viz. FD has a given ratio to E : and because the ratio of AB to
CD is given, as AB to CD, so make AG to A
CP ; therefore the ratio of AG to CF is given ;
and CF is given, wherefore (2. dat,) AG is
given: and because as AB to CD, so is AG G-
to CF, and so is (19. 5.) GB to FD ; the ratio P

of GB to FD is given. And the ratio of FD
to E is given, wherefore (9. dat.) the ratio^ of
GB to E is given, and AG is given ; therefore
GB, the excess of AB above a given magui- B D E

tude AG, has a given ratio to E.

Cor. 1. And if the first have a given ratio to the second, and
the excess of the first above a given magnitude have a given ratio
to the third ; the excess of the second above a given magnitude
shall have a given ratio to the third. For, if the second be called
the first, and the first the second, this corollary will be the same
with the proposition.

Cor. 2. Also, if the first have a given ratio to the second, and
the excess of the third above a given magnitude have also a given

» See Note.

EQQUD*S DATA.

307

ratio to the second, the same excess shcJl hare a given ratio to the
first ; as is evident from thd 9th dat.

PROP. XXV,

17,

If there be three magnitudes, the excess of the first whereof
above a given magnitude has a given ratio to the second; and
the excess of the third above a given magnitude has a given
ratio to the same second : the first shall either have a given ratio
to the third, or the excess of one of them above a given magni-
tude shall have a given ratio to the other.

Let AB, C, DE be three magnitudes, and let the excesses of
each of the two AB, DE above given magnitudes have given
ratios to C; AB, DE feither have a given ratio fo one another,
or the excess of one of them above a given magnitude has a given
ratio to the other.

Let FB, the excess of AB above a given magnitude AP, have
a given ratio to C; and let GE, the excess A
of DE above the given magnitude DG, have
a given ratio to C ; and because FB, GE have
each of them a given ratio to C, they have
a given ratio (9. dat.) to one another. But to
FB, GE the given magnitu*s AP, DG are
added; therefore (18. dat.) the whole magni-
tudes AB, DE have either a given ratio to one
another, or the excess of one of them above a
given magnitude has a given ratio to the other.

p- —

PROP. XXVI.

18.

If there be three magnitudes ; the excesses of one of which
above given magnitudes have given ratios to the other two mag-
nitudes ; these two shall either have a given ratio to one another,
or the excess of one of them above a given magnitude shall
have a given ratio to the other.

Let AB, CD, EP be three magnitudes, and let GD the excess
of one of them CD above the given magnitude CG have a given
ratio to AB ; and also let KD the excess of the same CD above the
given magnitude CK have a given ratio to EP : either AB has a
given ratio to EP, or the excess of one of them above a given mag-
nitude has a given ratio to the other.

Because GD has a given ratio to AB, as GD to AB, so make
CG to HA i therefore the ratio of CG to HA is given : and CQ is
given, wherefore (2. dat.) HA is givmx ; and because as GD to AB,
?J> If CG to HA. and so is (12. 5.) CD to HB; the raUo of CD to
HB is given : also bet^ause Kt> has a given ratio to EF, as KD to

308

EUCUD^fl DATA.

£F» 80 make CK to LE: therefore the ratio H
of CK to LE is gi^en; and CK is given,
wherefore LE (2. dat.) is given: and because
as KD to EF, so is CK to LE, and so (12. A-
6.) is CD to LP ; the ratio of CD to LF is
given: but the ratio of CD to HB is given,
wherefore (9. dat.) the ratio of HB to LF is
given: and from HB, LF the given magni-
tudes HA, LE being taken, the remainders .
AB, EF shall either have a given ratio to one
another, or the excess of one of them above a given magnitude
has a given ratio to the other (19. dat.).

Another Demonstration.

Let AB, C, DE be three magnitudes, and let the excesses of
one of them C above given magnitudes have given ratios to AB
and DE : either AB, DE have a given ratio to one another, or the
excess of one of them above a- given magnitude has a given ratio
to the other.

Because the excess of C above a given magnitude has a given
ratio to AB; therefore (14. dat.) AB together with a given mag-
nitude has a given ratio to C : let this given p
magnitude be AF, wherefore FB ha%e given
ratio to C : also because the excess of C above
a given magnitude has a given ratio to DE ;
therefore (14. dat.) DE together with a given
magnitude has a given ratio to C : let this
given magnitude be DG, wherefore GE has
a given ratio to C : and FB has a given ratio
to C, therefore (9. dat.) the ratio of FB to GE is given : and from
PB, GE the given magnitudes AF, DG being taken, the remain-
ders AB, DE either have a given ratio to one another, or the
excess of one of them above a given magnitude has a given ratio to
the other (19. dat.).

A—

PROP. XXVII.

19.

If there be three magnitudes, the excess of the first of which
above a given magnitude has a given ratio to the second ; and
the excess of the second above a given magnitude has also a
given ratio to the third ; the excess of the first above a given
magnitude shall have a given ratio to the third.

Let AB, CD, E be three magnitudes, the excess of the first of
which AB above the given magnitude AG, viz. GB, has a given
ratio to CD ; and FD the excess of CD above the given magnitude
OF, has a given ratio to E : the excess of AB above a given magni-
tude has a given ratio to R

Because the ratio of GB to CD is given, as GB to CD, so make

EUCLID 8 DATA.

309

G— C

H-

»—

P-^

GH to CP : therefore the ratio of GH to CF
is given ; and CP is given, wherefore (2. dat.)
GH is given : and AG is given, wherefore the
whole AH is given : and because as GB to
CD, so is GH to CF, and so is (19. 5.) the re-
mainder HB to the remainder FD; the ratio
of HB to FD is given : and the ratio of FD to
E is given, wherefore (9. dat.) the ratio of HB
to£ is given: and AH is given; therefore HB,
the excess of AB above a given magnitude AH, has a given -ratio
toE.

" Otherwise,

Let AB, C, D, be three magnitudes, the excess EB of the first
of which AB above the given magnitude AE has a given ratio to
C, and the excess of C above a given magni- .
tude has a given ratio to D : the excess of AB
above a given magnitude has a given ratio
toD.

Because EB has a given ratio to C, and the
excess of C above a given magnitude has a
given ratio to D ; therefore (24. dat.) the ex-
cess of EB above a given magnitude has a
given ratio to D: let this given magnitude
- be EF ; therefore FB, the excess of EB above EF, has a given
ratio to D : and AF is giyen, because AE, EF are given ; therefore
FB, the excess of AB above a given magnitude AF, has a given
ratio to D."

PROP. XXVIII. 25.

If two lines given in position cut one another, the point or
points in which they cut one another are given.*

Let two lines AB, GD given is position cut one another in the
point E ; the point E is given.

Because the lines AB, CD are
given in position, they have always
the same situation (4. def), and
therefore the point, or points, in
which they cut one another, have al-
ways the same situation : and because
the lines AB, CD can be found (4.
def.), the point, or points, in which
they cut one another, are likewise
found ; and therefore are given in po-
sition (4. def.) C D

PROP. XXIX. 26.

If the extremities of a straight line be given in position ; the
straight line is given in position and magnitude.
Because the extremities of the straight line are given, they can

* See Note.

310 Euclid's data.

be found (4. def.) : let these be the points A, B, between whicli
a straight line AB can be drawn (1. pos- . ^

tulate) ; this has an invariable position,

because between two given points there can be drawn but one
straight line : and when tiie straight line AB is drawn, its magnitude
is at the same time exhibited, or given : therefore the straight line
AB is given in position and magnitude.

PROP. XXX. 27.

fr one of the extremities of a straight line given in position
and magnitude be given: the other extremity snail also be given.

Let the point A be given, to wit, one of the extremities of a
straight line given in magnitude, and which lies in the straight line
AC given in position ; the other extremity is also given.

Because the straight line is given in magnitude, one equal to it
can be found (1. def.); let this be the straight line D: from the
greater straight line AC cut off AB equal . r p

to the lesser D : therefore the other ex- .

tremity B of the straight line AB is ''

found : and the point B has always the D
same situation; because any oi^er point

in AC, upon the same side of A, cuts off between it and the point
A a greater or less straight line than AB, that is, than D^; therefore
the point B is given (4. def.) : and it is plain another such point can
be found in AC, produced upon the other side of the point A.

PROP. XXXI. 28.

If a straight line be drawn through a given point parallel to a
straight line given in position ; that straight line is given in posi-
tion.

Let A be a given point, and BC a straight line given in position ;
the straight line drawn through a parallel to BC is given in po-
sition.

Through A draw (31. 1.) the straight ^ a v

line DAE parallel to BC; the straight ^ V^ ^

line DAE has always the same position, — — — |
because no other straight line can be B C

drawn through A parallel to BC; there-

fore the straight line DAE, which has been found, is given (4. def.)
in position.

PROP. XXXII. 29,

If a straight line be drawn to a given point in a straight line

ffiven in position, and makes a given angle with it ; that straight
ine is given in position.

Let AB be a straight line given in position, and C a given point

eoolid'b data. 811

Iri it ; the straight line drawn to O,
which makes a given angle with CB,
is given in position.

Because the angle is given, one
equal to it can be found (1. def) ; let
this be the angle at D : at the given A
point C, in the given straight line
AB, make (23. 1.) the angle ECB
equal to the angle at D : therefore
the straight line EG has always the
same situation, because any other
straight line FC, drawn to the point
C, makes with CB a greater or less angle than the angle ECB» or
the angle at D : therefore the straight line EC, which has l^en
found, is given in position.

It is to be observed, that there are two straight lines EC, GC
upon one side of AB that make equal angles with it, and which
make equal angles with it when produced to the other side.

PROP. xxxm. 30.

If a straight line be drawn from k given point to a straight
line given in position, and makea a given angle with it ; that
straight line is given in position.

From the given point A, let the straight line AD be drawn to the
straight line BC given in position, and make with it a given angle
ADC ; AD is given in position. E A F

Through the point A, draw (31. 1.) the
straight line EAF parallel to BC; and
because through the given point A, the

straight line EAF is drawn parallel to BC, n r*

which is given in position, EAF is there- DC

fore given in position (31. dat): and because the straight line AD
meets the parallels, BC, EF, the angle EAD is equal (29. 1.) to
the angle ADC ; and ADC is given, wherefore also the angle EAD
is given : therefore, because the straight line DA is drawn to the
given point A in the straight line EF given in position, and makes
with it a given angle EAD, AD is given (32. dat.) in position.

PROP. XXXIV. 31.

If from a given point to a straight line given in position, a
straight line be drawn which is given ifi magnitude ; the same
is also given in position.* ^

Let A be a given point, and BC a straight line given in position ;
a straight line given in magnitude drawn from the point A to BC is
given in position.

Because the straight line is given in magnitude, one equal to

♦See Note.

312 BUCLII>*S DATA.

it can be found (1. def.); let this be the straight line D : from the
point A draw A£ perpendicular to BC ; and be- A

cause AE is the shortest of all the straight lines
which can be drawn from the point A to BC, the
straight line D, to which one equal is to be

drawn from the point A to BC, cannot be less

than AE. If therefore D be equal to AE, AE b E

is the straight line given in magnitude, drawn ^^

from the given point A to BC : and it is evident

that AE is given in position, (33. dat.), because it is drawn from

the given point A to BC, which is given in position, and makes with

BC the given angle AEC. '

But if the straight line D be not equal to AE, it must be greater
than it: produce AE, and make AF equal to D; and from the
centre A, at the distance AF, describe the circle GFH, and join
AG, AH: because the circle GFH is given in position (6. def),
and the straight line BC is also given in .position ; therefore their
intersection G is given (28. dat.); A

and the point A is given ; where-
fore AG is given in position (29.
dat), that is, the straight line AG
given in magnitude, (for it is equal
to D) and drawn from the given
point A to the straight line BC F

given in position, is also given in D

position : and in like manner AH is given in position : therefore in
this case there are two straight lines AG, AH of the same magni-
tude, which can be drawn from a given point A to a straight Tine
BC given in position.

PROP. XXXV. 32.

If a straight line be drawn between two parallel straight
lines given in position, and makes giVen angles with them, the
straight line is given in magnitude.

Let the straight line EF be drawn between the parallels AB, CD,
which are given in position, and make the given angles BEF, EFD :
EF is given in magnitude.

In CD take the given point G, and through G draw (31. 1.) GH
parallel to EF : and because CD meets the parallels GH, EF, the
angle EFD is equal (29. 1.) to the angle a EH B

HGD : and EFD is a given angle ; where-
fore the angle HGD is gi^ftn ; and because
HG is drawn to the given point G, in the
straight Jine CD, given in position, and
makes a given angle HGD : the straight q "^ q ~^

line HG is given in position (32. dat.) : and

AB is given in position : therefore the point H is given (28. dat.), and
the point G is cdso given, wherefore GH is given in magnitude (29.
dat.) and EF is equal to it, therefore EF is given in magnitude.

BUCLID*S DATA. 313

PROP. XXXVI. 38.

If a straight line given in magnitude be drawn between two
parallel straight lines given in position, it shall make given an-
gles with the parallels.*

Let the straight line EP given in magnitude be drawn between

the parallel straight lines AB, CD, which are A E H B

given in position: the angles AEF, EFC shall
be given.

Because EP is giv^n in magnitude, a straight
line equal to it can be found (1. def.) : let this be
G: in AB take a given point H, and from it draw -;

(12. 1.) HK perpendicular to CD; therefore the ^ P K D

straight line G, that is, EP, cannot be Igss than G-

HK: and if G be equal to HK, EP also is equal to it : wherefore EP
is at right angles to CD : for if it be not, EP would be greater than
HK, which is absurd. Therefore the angle EPD is a right, and con-
sequently a given angle.

But if the straight line G be not equal to HK, it must be greater
than it : produce HK, and take HL, equal to G, and from the centre
H, at the distance HL, describe the circle MLN, and join HM, HN :
and because the circle (6. def.) MLN, and the straight line CD, are
given in position, the points M, N are (28. dat.) given : and the point
H is given, wherefore the straight A E H B

lines HM, HN, are given in posi-
tion (29. dat.) and CD is given
in position : therefore the angles
HMN, HNM, are given in posi-
tion (A. def) : of the straight lines C P

HM, HN, let HN be that which is G —

not parallel to EP, for EP cannot be parallel to both of them ; and
draw EO parallel to HN : EO therefore is equal (34. 1.) to HN, that
is to G ; and EP is equal to G, wherefore EO is equal to EP, and the
angle EPO to the angle EOP, that is, (29. 1.) to the given angle
HNM ; and because the angle HNM, which is equal to the angle
EPO, or EPD, has been found : therefore the angle EPD, that is, the
angle AEP, is given in magnitude (1. def.); and consequently the
angle EPC.

PROP. XXXVII. K.

Ip a straight line given in magnitude be drawn from a point to
a straight line given in position, in a given angle ; the straight
line drawn through that point parallel to the straight line given
in position, is given in position."*^

• See Note.
40

314 euoud's data.

Let the straight line AD given in magnitude be drawn from the
point A to the straight line BC, given in posi- E A H F

tion, in the given angle ADC: the straight
line EAF drawn through A parallel to BC is
given in position.

In BC take the given point G, and draw

GH parallel to AD : and because HG is drawn B D G • C

to a given point G in the straight line BC given in position, in a given
angle HGC, for it is equal (29. 1.) to the given angle ADC; HG is
given in position (32. dat.) ; but it is given also in magnitude, be*
cause it is equal to (34. I.) AD which is given in magnitude; there-
fore because G, one of the extremities of the straight line GH, given
in position and magnitude is given, the other extremity H is given
(30. dat.) ; and the straight line EAF, which is drawn through the
given point H parallel to BC given in position, is therefore given (31.
dat.) in position.

PROP. XXXVIII. 34.

If a straight line be drawn from a given point to two parallel
straight lines given in position, the ratio of the segments between
the given point and the parallels shall be given.

Let the straight line EFG be drawn from the given point E to the
parallels AB, CD ; the ratio of DF to EG is given.

From the point E draw EHK perpendicular to CD ; and because
from a given point E the straight line EK is drawn to CD which is
given in position, in a given angle EEC ; EE is given in position

F H B

CGK DC KGD

(33. dat.) ; and AB, CD are given in position : therefore (28. dat.) the
points H, K are given ; and the point E is given ; wherefore (29. dat.)
EH, EK are given in magnitude, and the ratio (1. dat.) of them is
therefore given. But as EH to EK, so is EF to EG, because AB,
CD are parallels ; therefore the ratio of EF to EG is given.

PROP. XXXIX. 35, 36.

If the ratio of the segments of a straight line between a given
point in it and two parallel straight lines be given, if one of the
parallels be given in position, the other is also given in position.

BUC1.ID ft DATA.

315

From the given point A, let the straight line AED be drawn to the
two parallel straight lines FG, BC, and let the ratio of the segments
AE, AD be given ; if one of the parallels BC be given in position, the
other FG is also given in position.

From the point A, draw AH perpendicular to BC, and let it meet
FG in K: and because AH is drawn from the given point A to
the straight Ime BC given in position, and makes a given angle

E K

B D H C B

AHD; AH is given (33. dat.) in posi-
tion; and BC is likewise given in posi-
tion ; therefore the point H is given (29.
dat.): the point A is also given; where-
fore AH is given in magnitude (29. dat.) ;
and because FG, BC are parallels, as AE
to AD, so is AK to AH ; and the ratio of
AE to AD is given, wherefore the ratio of
AK to AH is given ; but AH is given in magnitude, therefore (2.
dat.) AK is given in magnitude ; and it is also given in position, and
the point A is given ; wherefore (30. dat.) the point K is given. And
because the straight line FG is drawn through the given point K
parallel to BC which is given in position, therefore (31. dat.) FG is
given in position.

-7

— ,

■^

PROP. XL.

37, 38.

Ir the ratio of the segments of a straight line into which it is
cut by three parallel straight lines, be given; if two of the paral-
lels are given in position, the third is also given in position.*

Let AB, CD, HK be three parallel straight lines, of which AB, CD
are given in position ; and let the ratio of the segments GE, GF into
which the straight line GEF is cut by the three parallels, be given ;
the third parallel HK is given in position.

In AB take a given point L, and draw LM perpendicular to
CD, meeting HK in N; because LM is drawn from the given
point L to CD which is given in position, and makes a given
angle LMD ; LM is given in position (33. dat.) ; and CD is given
in position, wherefore the point M is gixen (28. dat.) ; and the point
L is given; LM is therefore given in magnitude (20. dat.): and
because the ratio of GE to GF is given, and as GE to GF, so i»

« See Note.

316

Euclid's DjkTA.

G N

CP' M D CP M D

NL to NM ; the ratio of NL to NM is given ; and therefore (cor. 6.
or 7. dat.) the ratio of ML to LN is given ; but LM is given in mag-
nitude (cor. 6. or 7. dat.), wherefore (2. dat.) LN is given in mag-
nitude ; and it is also given in position, and the point L is given,
wherefore (30. dat.) the point N is given ; and because the straight
line HK is drawn through the given point N parallel to CD which is
given in position, therefore HK is given in position (31. dat.).

PROP. XLL

Ir a straight line meets three parallel straight lines which are
given in position, the segments into which they cut it have a
given ratio.

Let the parallel straight lines AB, CD, EF, given in position, be
cut by the straight line GHK ; the ratio* of GH to HK is given.

In AB take a given point L, and A G L B

draw lM perpendicular to CD, meet-
ing EF in N ; therefore (33. dat.) LM
is given in position; and CD, EF are
given in position, wherefore the points
M, N are given; and the point L is
given ; therefore (29. dat.) the straight
lines LM, MN are given in magnitude ; E K N ^ P

and the ratio of LM to MN is therefore given (1. dat.) : but as LM
to MN, so is GH to HK ; wherefore the ratio of GH to HK is given.

PROP. XLIL

39.

If each of the sides of a triangle be given in magnitude, the
triangle is given in species.

Let each of the sides of the triangle ABC be given in magnitude,
the triangle ABC is given in species.

Make a triangle (22. 1.) A D

DEP, the sides of which are
equal, each to each, to the
given straight lines AB, BC,
GA, which can be done; be-
cause any two of them must
be greater than the third; and B C ®, . ». ^

let DE be equal to AB, EF to BC, and FD to CA'; .and' because
the two sides ED, DP are equal to the two BA, AC, each to each,
and the base EF equal to the base BC ; the angle EDF is equal

EUOLID'S DATA.

317

(8. 1.) to the angle BAG; therefore, because the angle EDP which
is equal to the angle BAG, has been found, the angle BAG is given
(1. def ) ; in like manner the angles at B, G are given. And because
the sides AB, BC, GA are given, their ratios to one another are given
(1. dat.), therefore the triangle ABG is given (3. def ) in species.

PROP. XLin.

40.

. If each of the angles of a triangle be given in magnitude, the
triangle is given in species.

Let each of the angles of the triangle ABC be given in» magnitude,
the triangle ABC is given in species.

llake a straight line DE given in po- A

sition and magnitude, and at the points
D, E make (23. 1.) the angle EDF equal
to the angle BAG ; and the angle DEF
equal to ABG ; therefore the other an-
gles EFD, BGA are equal, and each of
the angles at the points A, B, C is given ; B G E F

wherefore each of those at the points D, E, F is given : and because
the straight line FD is drawn to the given point D in DE, which is
given in position, making the given angle EDF ; therefore DF is
given in position (32. dat.). In like manner EF also is given in po-
sition ; wherefore the point F is given : and the points D, E are
given; therefore each of the straight lines DE, EF, FD is given (29.
dat.) in magnitude; wherefore the triangle DEF is given in species
(42. dat.) ; and it is similar (4. 6. 1 . def. 6.) to the triangle ABG :
which is therefore given in species.

PROP. XLIV.

41.

If one of the triangles of a triangle be given, and if the sides
about it have a given ratio to one another ; the triangle is given
in species.

Let the triangle ABG have one of its angles BAG given, and let
the sides BA, AC about it have a given ratio to one another ; the
triangle ABG is given in species.

Take a straight line DE given in position and magnitude, and
at the point D, in the given straight line DE, make the angle
EDF equal to the given angle BAG; wherefore the an'gle EDP
is given ; and because the straight line FD is drawn to the given
point D in ED which is given in position, making the given
angle EDF ; therefore FD is given in A

position (32. dat.). And because the
ratio of BA to AC is given, make the
ratio of ED to DF the same with it,
and join EF; and because the ratio of
ED to DP is given, and ED is given,

therefore (2. dat.) DF is given in mag- B C E

nitttde : and it is given also in position, and the point D is given,

318

^DGIilD'S DATA.

wherefore the point F is given (30. dat.) ; and the points D, B are
given, wherefore DE, EF, FD are given (29. dat.) in magnitude ; *nd
the triangle DEF is therefore given (42. dat.) in species ; and because
the triangles ABC, DEF have one angle BAG equal to one angle
£DF, and the sides about these angles proportionals ; the triangles
are (6. 6,) similar ; but the triangle DEF is given in species, and
therefore also the triangle ABC.

PROP. XLV.

42.

DEF

If the sides of a triangle have to one another given ratios, the
triangle is given in species.

Let the sides of the triangle ABC have given ratios to one another,
the triangle ABC is given in species.

Take a straight line D given in magnitude; and because the
ratio of AB to BC is given, make the ratio of D to E the same
with (it ; and D is given, therefore (2. dat.) E is given. And be-
cause the ratio of BC to CA is given, to this make the ratio of E
to F the same ; and E is given, and therefore (2. dat.) F ; and be*
cause as AB to BC, so is D to E; by composition AB and BC
together are to BC, as D and E to
P; but as BC to CA, so is E to F;
therefore, ex ssquali, (22. 5,) as AB
and BC are to CA, so are D and E
to F, and AB and BC are greater
(20. 1.) than CA ; therefore D and
Eare greater (A. 5.) than F. In
the same manner any two of the
three D, E, F are greater than the
third. Make (22. 1 .) the triangle GHK
whose sides are equal to D, E, F, so
that GH be equal to D, HK to E, and
KG to F ; and because D, E, F, are
each of them given, therefore GH,
HK^ KG are each of them given in magnitude ; therefore the triangle
GHK is given (42. dat.) in species ; but as AB to BC, so is (D to E,
that is) GH to HK ; and as BC to CA, so is (E to F, that is) HK to
KG ; therefore, ex asgualU as AD to AC, so is GH to GK. Where-
fore (5. 6.) the triangle ABC is equiangular and similar to the tri-
angle GHK ; and the triangle GHK is given in species ; therefore
also the triangle ABC is given in species.

CoR. If a triangle is required to be made ; the sides of which
shall have the same ratios which three given straight lines D, E, F
have to one another ; it is necessary that every two of them be
greater than the third.

PROP. XLVI. 43.

If the sides of a right angled triangle about one of the acute
angles have k given ratio to one another; the tmangle is given
in species.

BVCIilo'B DATA. SIO

Let the aides AB« BC about the acute angle ABC of the triangle
ABC, which has a right angle at A, have a given ratio to one ano«
tber ; the triangle ABC is given in species.

Take a straight line DE given in position and magnitude ; and
because the ratio of AB to BC is given, make as AB to BC, so
DE to EF ; and because DE has a given ratio to EF, and DE ia
given, therefore (2. dat.) EF is given ; and because as AB to BC,
so is DE to EF; and AB is less (19. 1.) than BC, therefore DE is
less (a, 5.) than EF. From the point D draw DG at right angles
to DE, and from the centre E, at
the distance EF, describe a circle
which shall meet DG in two
points; let G be either of them,
and join EG; therefore the cir-
cumference of the circle is given
(6. def.) in position; and the
straight line DG is given (35J.
dat.) in position, because it is drawn to the given point D in DE
given in position, in a given angle ; therefore (28. dat) the point
G is given ; and the points D, E are given : wherefore DE, B3Q,
GD are given (29. dat.) in magnitude, and the triangle DEG in
species (42. dat.). And because the triangles ABC, DEG have
the angle BAG, equal to the angle EDG, and the sides about the
angles ABC, DEG proportionals, and each of the other angles
BOA, EGD less than a right angle; the triangle ABC is eqai*
angular (7. 6.) and similar to the triangle DEG : but DEG is given
in species ; therefore the triangle ABC is given in species : and, in
the same manner, the triangle made by drawing a straight line
from E to the other point in which the circle meets DG is given
in species.

PROP. XLVU. 44.

If a triangle has one of its angles which is not a right angle
given, and if the sides about another angle have a given ratio to
one another ; the triangle is given in species.

Let the triangle ABC have one of its angles ABC a given but
not a right angle, and let the sides BA, AC about another angle
BAC have a given ratio to one another ; the triangle ABC Is given
in species.

First, let the given ratio be the ratio of A

equality, that is, let the sides BA, AC, and con-
sequently the angles ABC, ACB be equal; and
because the angle ABC is given, the angle ACB,
and also the remaining (32. 1.) angle BAC is
given; therefore the triangle ABC is given (43.
dat.) in species ; and it is evident that in this
case tlie given angle ABC must be acute.

Next, let the given ratio be the ratio of a less to a greater,
that is, let the side AB adjacent to the given angle be less than

820

EUCLID S DATAk

:::i.P

the side AC ; take a straight line DE glYen in position and magnitude^
and make the angle DEF equal to the given angle ABC ; therefore
EF is given (32. dat.) in position^ and because the ratio of BA to AC
is given, as BA to AC, so make ED to A
DG ; and because the ratio of ED to DG
is given, and ED is given, the straight line
DG is given (2. dat.)* and BA is less than
AC, therefore ED is less (A. 5.) than DG. B"
From the centre D at the distance DG de-
scribe the circle GF meeting EF in F, and
join DF ; and because the circle is given
(6. def.) in position, as also the straight
line EF, the point F is given (28. dat.) ;
and the points D, E are given ; wherefore
the straight lines DE, EF, FD are given
(29. dat.) in magnitude, and the triangle
DEF in species (42. dat). And because G
BA ii^less than AC, the angle ACB is less (18. 1.) than the angle
ABC, and therefore ABC is less (1. 7. 1.) than a right angle. In
the same manner, because ED is less than DG or DF, the angle
DFE is less than a right angle: and because the triangles ABC,
DEF have the angle ABC equal to the angle DEF, and the sides
about the angles BAC, EDF proportionals, and each of the other
angles ACB, DFE less than a right angle ; the triangles ABC, DEF
are (7. 6.) similar, and DEF is given in species, wherefore the
triangle ABC is also given in species.

Thirdly, let the given ratio be the ratio of a greater to a less,
that is, let the side AB adjacent to the given angle be greater
than AC ; and, as in the last case, take a
straight line DE given in position and A

magnitude, and make the angle DEF equal
to the given angle ABC; therefore EF is
given (32. dat.) in position: also draw
DG perpendicular to EF; therefore if the
ratio of BA to AC be the same with the
ratio of ED to the perpendicular DG,
the triangles ABC, DEG are similar
(7. 6.), because \he angles ABC, DEG
are equal, and DGE is a right angle:
therefore the angle ACB is a right angle,
and the triangle ABC is given in (43. dat.)
species.

But if, in this last case, the given ratio of BA to AC be not
the same with the ratio of ED to DG, that is, with the ratio
of BA to the perpendicular AM drawn from A to BC ; the
ratio of BA to AC must be less thati (8. 5.) the ratio of BA to
AM, because AC is greater than AM. Make as BA to AC, so
ED to DH ; therefore the ratio of ED to DH is less than the ratio
of (BA to AM, that is, than the ratio of) ED to DG ; and conse>
quently DH is greater (10. 5.)tthan DG; and because BA is great-

Euclid's data.

er than AC, ED is greater (A. 5.) than DH.
From the centre D, at the distance DH, describe
the circle KHF which necessarily meets the
straight line EF in two points, because DH is
greater than DG, and less than DR Let the
circle meet EF in the points F, K which are
given, as was shown in the preceding case ; and
DF, DK being joined, the triangles DEF, DEK
are given in species, as was there shown. From
the centre A, at the distance AC, describe a cir-
cle meeting BC again in L : and if the angle ACB E K
be less than a right angle, ALB must be greater
than a right angle ; and on the contrary. In the
same manner, if the angle DGF be less, than a right angle, DKE must
be greater than one ; and on the contrary. Let each of the angles
ACB, DFE be either less or greater than a right A

angle ; and because in the triangles ABC, DEF,
the angles ABC, DEF are equal, and the sides
BA, AC and ED, DF about two of the other an-
gles proportionals, the triangle ABC is similar
(7. 6.) to the triangle DEF. In the same man-
ner, the triangle ABL is similar to DEK. And
the triangles DEF, 'DEE are given in species ;
therefore also the triangles ABC, ABL are given
in species. And from this it is evident, that in
this third case there are always two triangles of « ^
a different species, to which the things mention-
ed as given in the proposition can agree.

PROP. XLVIII.

45.

If a triangle has one angle given, and if both the sides together
about that angle have a given ratio to the remaining side ; the
triangle is given in species.

Let the triangle ABC have the angle BAC given, and let the sides
BA, AC together about that angle have a given ratio to BC ; the tri-
angle ABC is given in species.

Bisect (9. 1.) the angle BAC by the straight line AD; therefore the
angle BAD is given. And because as BA to AC, so is (3. 6.) BD to
DC ; by permutation, as AB to BD, so is AC to
CD : and as BA and AC together to BC, so is
(12. 5.) AB to BD. But the ratio of BA and
AC together to BC is given, wherefore the ratio
of AB to BD is given, and the angle BAD is
given : therefore (47. dat.) the triangle ABD is °
given in species, and the angle ABD is therefore given : the angle
BAC is also given : wherefore the triangle ABC is given in species
(43. dat.).

BUCLID*8 DATA.

A triangle which shall have the things that are mentioned in the
proposition to be given, ca^ be found in the following manner. Let
EFQ be the given angle, and let the ratio of H to K be the given
ratio which the two sides about the angle EFG must have to the
third side of the triangle : therefore, because two sides of a triangle
are greater than the third side, the ratio of H to K must be the ratio
of a greater to a less. Bisect (9. I.) the angle EFG by the straight
line FL, and by the 47th proposition find a triangle of which EFL is
one of the angles, and in which the ratio of the sides about the angle
opposite to FL is the same with the ratio of H to K : to do which
take FE given in position and magnitude, and draw EL perpendicu-
lar to FL ; then, if the ratio of H to K be the same with the ratio of
FE to EL, produce EL, and let it meet FG in P : the triangle FEP is
that which was to be found : For it has the given angle EFG ; and
because this angle is bisected by FL, the sides EF, FP together are
to EP, as (3. 6.) FE to EL, that is, as H to K.

But if the ratio of H to K be not the same with the ratio of FE to
EL, it must be less than it, as was shown in prop. 47, and in this
case there are two triangles, each of
which has the given angle EFL, and the
ratio of the sides about the angle oppo-
site to FL the same with the ratio of
H to K. By prop. 47, find these trian-
gles EFMy EFN, each of which has the
angle EFL for one of its angles, and
the ratio of the side FE to EM or EN
the same with the ratio of H to K ; and let the angle EMF be greater,
and ENF less than a right angle. And because H is greater than
K, EF is greater than EN, and therefore the angle EFN, that is, the
angle NFG, is less (18. L) than the angle ENF. To each of these
add the angles NEF, EFN: therefore the angles NEF, EFG are less
than the angles NEF, EFN, FNE, that is, than two right angles :
tlierefore the straight lines EN, FG must meet together when pro-
duced ; let them meet in O, and produce EM to G. Each of the tri-
angles EFG, EFO has the things mentioned to be given in the pro-
position : for each of them has the given angle EFG ; and because
this angle is bisected by the straight line FMN, the sides EF, FG to-
gether have to EG the third side the ratio of FE to EM, that is, of H
to K. In like manner, the sides EF, FO together have to EO the
xatk) which H has to K.

PROP. XLIX. 46.

If a triangle has one angle given, and if the sides about an-
other angle, both together, have a given ratio to the third side;
the triangle is given in species.

Let the triangle ABC have one angle ABC given, and let the two
«ides BA, AC about another angle BAC have a given ratio to BC ;
the triangle ABC is given in species.

" ^ -

A>^-

/ >r >v^

l^Y^^

E N

Euclid's data.

Suppose the angle BAC to be bisected by the straight Ifoe AD :
BA and AC together are to BC, as AB^ BD, as was shown in the
preceding proposition. But the ratio of BA and AC together to BC
is given, therefore also the ratio of AB to BD is given. And the
angle ABD is given, wherefore (44. dat.) the triangle ABD is given
in species: and consequently the angle BAD, and its double the
angle BAC are gfven ; and the angle ABC A

is given. Therefore the triangle ABC is
given in species (43. dat).

A triangle which shall have the things
mentioned in the proposition to be given,
may be thus fbund. Let EFG be the
given angle, and the ratio of H to K the
given ratio: and by prop. 44, find the
triangle EFL, which has the angle EFG
for one of its angles, and the ratio of the
sides EF, FL about this angle the same
with the ratio of H to K ; and make the F L Q

angle LEM equal to the angle FEL. And because the ratio of H to
E is the ratio which two sides of a triangle have to the third, H
must be greater than K ; and because EF is to FL, as H to E,
therefore EF is greater than FL, and the angle FEL^ that is, LEM,
is therefore less than the angle ELF. Wherefore the angles LFE,
FEM are less than two right angles, as was shown in the foregoing
proposition, and the straight lines FL, EM must meet, if produced :
let them meet in G, EFG is the triangle which was to be found ; for
EFG is one of its angles, and because the angle EFG is bisected by
EL, the two sides FE, EG together have to the third side FG the
ratio of EF to FL, that is, the given ratio of H to E.

PROP. L. 76.

If from the vertex of a triangle, given in species, a straight
line be drawn to the base in a given angle, it shall have a given
ratio to the base.

From the vertex A of the. triangle ABC which is given in species,
let AD be drawn to the base BC in a given angle ADB; the ratio of
AD to BC is given.

Because the triangle ABC is given in speces,
the angle ABD is given, and the angle ADB is
given, therefore the triangle ABD is given (43.
dat.) in species ; wherefore the ratio of AD to
AB is given. And the ratio of AB to BC is
given ; and therefore (9. dat) the ratio of AD to
BC is given.

PROP. U. 47.

Rectilineal figures, given in species^ ar^e divided into trian-
gles which are given in species.

U4 £UCLID*8 DATA.

Let the rectilineal figure ABODE be given in species; *ABCDE may
be divided into triangles gifpn in species.

Join BE, BD ; and because ABODE is given in species, the angle
BAE is given (3. def ), and the ratio of BA to
AE is given (3. def.) ; wherefore the triangle
BAE is given in species (44. dat.)i and the
angle AEB is therefore given (3. def). But
the whole angle AED is given, and therefore
the remaining angle BED is given, and the
ratio of AE to EB is given, as also the ratio
of AE to ED ; therefore the ratio of BE to ED C D

is given (9. dat.). And the angle BED is given, wherefore the tri-
angle BED is given (44. dat.) in species. In the same manner, the
triangle BDO is given in species : therefore rectilineal figures wliicb
are given in species are divided into triangles given in species.

PROP. Ln. 48.

If two triangles given in species be described upon the satae
straight line, they shall have a given ratio to one another*

Let the triangles ABC, ABD, given in species, be described upon
the same straight line AB ; the ratio of the triangle ABC to the tri-
angle ABD is given.

Through the point C draw OE parallel to AB, and let it meet
DA produced in E, and join BE. Because the triangle ABC is
given in species, the angle BAO, that is, the angle ACE, is given ;
and because the triangle ABD is given in species, the angle DAft
that is, the angle AEO E

is given. Therefore the \<c;;[ ^ L

triangle ACE is given in

species ; wherefore the

ratio of EA to AC is

given (3. def.), and the

ratio of CA to AB is

given, as also the ratio of ^^ K

BA to AD ; therefore the ^

ratio of (9. dat.) EA to AD is given, and the triangle ACB is equal

(37. L) to the triangle AEB, and as the triangle AEB, or ACB, is to

the triangle ADB, so is (L 6.) the straight line EA to AD. But the

ratio of EA to AD is given, therefore the ratio of the triangle AOB

to the triangle ADB is given.

PROBLEM.

To find the ratio of two triangles ABC, ABD given in species, and
which are described upon the same straight line AB.

Take a straight line PG given in position and magnitude, and
because the angles of the triangles ABO, ABD are ^ven, at the
points P, G of the straight line FG, make the angles GPH, GFK
(28. 1.) equal to the angles BAG, BAD : and the angles FGH,

BJP

EUCLID 8 DATA.

FGK equal to the angles ABC, ABD, each fo each. Therefore
the triangles ABC, ABD are equiangular to the triangles FGH.
FGK, each to each. Through the point H draw HL paraUel to
FG, meeting KF produced in L. And because the angles BAC,
BAD are equal to the angles GFH, GFK, each to each ; therefore
the angles ACE, AE(y are equal to FHL, FLH, each to each, and
the triangle AEC equiangular to the triangle FLH. Therefor^
as EA to AC, so is LF to FH ; and as CA to AB, so HF to FG ;
and as BA to AD, so is GF to FK ; wherefore, ex sequali^ as EA
to AD, so is LF to FK. But, as was shown, the triangle ABC is
to the triangle ABD, as the straight line EA to AD, that is, as
LF to FK The ratio therefore of LF to FK has been found,
which is the same with the ratio of the triangle ABC to the tri-
angle ABD.

PROP LUI.

49.

If two rectilineal figures given in species be described upon the
same straight line, they shall have a given ratio to one another.*

Let any two rectilineal figures ABCDE, ABFG, which are given
in species, be described upon the same straight line AB ; the ratio
of them to one another is given.

Join AC, AD, AF: each of the triangles AED, ADC, ACB,
AGF, ABF is given (51. dat.) in species. And becatise the trian-
gles ADE, ADC given in species are D
described upon the same straight line
AD, the ratio of EAD to DAC is
given (52. dat.) ; and, by composition,
the ratio of EACD to DAC is given
(7. dat.). And the ratio of DAC to
CAB is given (52. dat.) because they
are described upon the same straight
line AC ; therefore the ratio of EACD
to ACB is given (9. dat.): and, by
composition, the ratio of ABCDE to H-
ABC is given. In the same man-
ner, ^the ratio of ABFG to ABF is given. But the, ratio of the
triangle ABC to the triangle ABF is given; wherefore (52. dat.),
because the ratio of ABCDE to ABC is given, as also the ratio
of ABC to ABF, and the ratio of ABF to ABFG ; the ratio of the
rectilineal ABCDE to the rectilineal ABFG is given (9. dat.).

PROBLEM.

To find the ratio of two rectilineal figures given in species, and
described upon the same straight line.

Let ABCDE, ABFG be two rectilineal figures given in species,
and described upon the same straight line AB, and join AC, AD,
AF. Take a straight line HK given in position and magnitude,
and by the 52d dat. find the ratio of the triangle ADE to the tri-

K L M N

* See Note.

326 Euclid's data.

angle ADC, and make the ratio of HK to KL the same with it
Find also the ratio of the triangle ABD to the triangle ACB.
And make the ratio of KL to LM the same. Also, find the ratio of
the triangle ABC to the triangle ABF and make the ratio of
LM to MN the same. And, lastly, find the ratio of the triangle
AFB to the triangle AFG, and make D

the ratio of MN to NO the same.
Then the ratio of ABCDE to ABFG
is the same with the ratio of HM to
MO.

Because the triangle EAD is to
the triangle DAC as the straight
line HK to KL; and as the triangle
DAC to CAB, so is the straight line
KL to LM ; therefore, by using com- K L M N

position as often as the number of H 1 — | — | — |— O

triangles requires, the rectilineal

ABCDE is to the triangle ABC, as the straight line HM to ML.
In like manner, because the triangle GAF is to FAB, as ON 'to
NM, by composition, the rectilineal ABFG is to the triangle
ABF, as MO to NM ; and, by inversion, as ABF to ABFG, so is
NM to MO. And the triangle ABC is to ABF, as LM to MN.
Wherefore, because as ABCDE to ABC, so is HM to ML; and as
ABC to ABF, so is LM to MN ; andas ABF to ABFG, so is MN to
MO : ex ssquali, as the rectilineal ABCDE to ABFG, so is the
straight line HM to MO.

PROP. LIV. 50.

If two straight lines have a given ratio to one another, the
similar rectilineal figures described upon them similarly, shall
have a given ratio to one another.

Let the straight lines AB, CD have a given ratio to one another,
and let the similar and similarly placed rectilineal figures E, F be
described upon them ; the ratio of E to F is given.

To AB,CD, let G be a third proportional:
therefore, as AB to CD, so is CD to G. And
the ratio of AB to CD is given, wherefore
the ratio of CD to G is given ; and conse-
quently the ratio of AB to G is also given
(9. dat). But as AB to G, so is the figure
K to the figure (2. cor. 20. 6.) F. There-
fore the ratio of E to F is given.

PROBLEM.

To find the ratio of two similar rectilineal figures, E, F, simOarly
described upon straight lines AB, CD which have a given ratio to
one another: let G be a third proportional to AB, CD.

Take a straight line H given in magnitude ; and because the
ratio of AB to CD is given, make the ratio of H to K the same

'Va

B C ]
H K L

Euclid's data. 327

with it ; and because H is given, K is given. AS H is to K, so make
K to L ; then the ratio of £ to F is the same with the ratio of H to
L : for AB is to CD, .as H to K, wherefore CD is to G, as K to L :
and, ex sequali, as AB to G so is H to L : but the figure E is to (2»
cor. 20. 6.) the figure F, as AB to G, that is, as H to L.

PROP, LV. 51.

Ip two straight lines have a given ratio to one another ; the
rectilineal figures given in species described upon them, shall
have to one another a given ratio.

Let AB, CD be two straight lines which have a given ratio to one
another: the rectilineal figures E, F given in species and described
upon them have a given ratio to one another.

Upon the straight line AB, describe the figure AG similar and
similarly placed to the figure F ; and because F is given in species,
AG is also given in species : therefore,
since the figures E, AG, which are
given in species, are described upon A
the same straight line AB, the ratio of
E to AG is given (53. dat), and be-
cause the ratio of AB to CD is given,
and upon them are described the si- H-
milar and similarly placed rectilineal

figures AG, F, the ratio of AG to F is given (54. dat.) : and the ratio
of AG to E is given: therefore the ratio of E to F is given (9. dat.).

PROBLEM.

To find the ratio of two rectilineal figures E, F given in speeies,
and described upon the straight lines AB, CD which have a given
ratio to one another.

Take a straight line H given in magnitude ; and because the rec-
tilineal figures E, AG given in species are described upon the same
straight line AB, find their ratio by the 53d dat. and make the ratio
of H to K the same : K is therefore given ; and because the similar
rectilineal figures AG, F are described upon the straight lines AB, CD,
which have a given ratio, find their ratio by the 54th dat. and make
the ratio of K to L the same : the figure E has to F the same ratio
which H has to L : for by the construction, as E is to AG, so is H to
K; and as AG to F, i^o is K to L ; therefore, ex aequalU as E to F, so
is H to L.

PROP. LVI. 52.

Ir a rectilineal figure given in species be described upon a
straight line given in magnitude, the figure is given in magnitude.

Let the rectilineal figure ABCDE given in species be described

828

Euclid's data.

upon the straight line AB given in magnitj^de; the figure ABCDE is
given in magnitude.

Upon AB let the square AF be described ; C .

therefore AF is given in species and magni-
tude, and because the rectilineal figures
ABCDE, AF given in species are described D
upon the same straight line AB, the ratio of
ABCDE to AF is given (53. dat.) : but the
square AF is given in magnitude, therefore
i'Z. dat.) also the figure ABCDE is given in
magnitude.

PROBLEM.

To find the magnitude of a rectilineal figure
given in species described upon a straight line
given in magnitude.

Take the straight line GH equal to the given
straight line AB, and by the 53d dat. find the
ratio which the square AF upon AB has to the ^
figure ABCDE ; and make the ratio of GH to HK the same ; and
upon GH describe the square GL, and complete the parallelogram
LHKM ; the figure ABCDE is equal to LHKM : because AF is to
ABCDE, as the straight line GH to HE, that is, as the figure GL to
HM ; and AF is equal to GL ; therefore ABCDE is equal to HM
(14. 5.).

PROP. LVII.

53.

Ir two rectilineal figures are given in species, and if a side of
one .of them has a given ratio to a side of the other ; the ratios of
the remaining sides to the remaining sides shall be given.

Let AC, DF, be two rectilineal figures given in species, and let
the ratio of the side AB to the side DE be given, the ratios of the
remaining sides to the remaining sides are also given.

Because the ratio of AB to DE is given, as also (3. def.) the ratios
of AB to BC, and of DE to EF, the ratio of BO to EF is given (10.
dat). In the same manner, the ratios of the
other sides to the other sides are given.

The ratio which BC has to EF may be
found thus ; take a straight line G given in
magnitude, and because the ratio of BC to
BA is given, make the ratio of G to H the
same ; and because the ratio of AB to DE
is given, make the ratio of H to K the same ;
and make the ratio of K to L the same with
the given ratio of DE to EF. Since there-
fore as BC to BA, so is G to H ; and as BA
to DE, so is H to K; and as DE to EF, so is

K to L ; ex sequali, BC is to EF, as G to L ; therefore the ratio of G
to L has been found, which is the same with the ratio of BC to EF.

G H K L

H7CLH)*8 DATA. 8t9

PROP. LVffl. G.

If two similar rectilineal figures have a given ratio to one ano-
ther, their homologous sides have also a given ratio to one another.*

L^t the two similar rectilineal figures, A, B, have a given ratio to
one another, their homologous sides have also a given ratio.

Iiet the side CD be homologous to £F, and to CD, £F let the
straight line G be a third proportional. As therefore (2. cor. 20. 6.)
CD to G, so is the figure A to B ; and the
ratio of A to B is given, therefore the ratio
of CD to G is given ; and CD, EF, G are r j:^ , ^

proportionals ; wherefore (13. dat) the ratio / I / -^

of CD to EF is given. C D £ F G

The ratio of CD to EF may be found

thus : Take a straight line H given in „ i xr

magnitude ; and because the ratio of the !-• K.

figure A to B is given, make the ratio of H to K the same with it ;
And, as the 13th dat. directs to be done, find a mean proportional
L between H and K ; the ratio of CD to EF is the same with that
of H to L. Let G be a third proportional to CD, EF : therefore as
CD to G, so is (A to B, and so is) H to K ; and as CD to EF, so 19
H to L, as is shown in the 1 3th dat.

PROP. UX, 54.

Ir two rectilineal figures given in species have a given ratio
to one another, their sides shall likewise have given ratios to one
another.*

Let the two rectilineal figures A, B, given in species, have a given ratio
to one another, their sides shall also have given ratios to one another.

If the figure A be similar to B, their homologous sides shall have
a given ratio to one another, by the preceding proposition ; and be-
cause the figures are given in species, the sides of each of them have
given ratios (3. def.) to one another ; therefore each side of two of
tfaeni has (9. dat) to each side of the other a given ratio.

But if the figure A be not similar to B, let CD, EJF b^ any two
of their sides; and upon EF conceive the figure EG to be described
similar and similarly placed to the
figure A, so that CD, EF be homo-
logous sides; therefore EG is given
in species ; and the figure B is given
in species ; wherefore (53. dat.) the
ratio of B to EG is given ; and the
ratio of A to B is given, therefore

(9. d^t.) the ratio of the figure A to H ^ — —

EG \& given; and A is similar to K

EG ; therefore (58. dat.) the ratio M r—

of the §ide CD to EF is given ; and L — « — >■ ■

coosBcjuently (9. dat.) the ratios of

the remaining sides to the remaining sides are given.

« Site Note.
42

330

enCLID 8 DATA.

The ratio of CD to EF may be found thus : take a straight line H
given in magnitude, and because the ratio of the figure A to B is
given, make the ratio of H to K the same with it. And by the 53d
dat. find the ratio of the figure B to EG, and make the ratio of K to
L the same : Between H and L find a mean proportional M ; the
ratio of CD to EF is the same with the ratio of H to M ; because
the figure A is to B as H to K ; and as B to EG, so is E to L ; ex
aiquali, as A to EG so is H to L : and the figures A, EG are similar,
and M is a mean proportional between H and L ; therefore, as was
shown in the preceding proposition, CD is to EF as H to M.

PROP. LX.

55.

Ip a rectilineal figure be given in species and magnitude, the
sides of it shall be given in magnitude.

Let the rectilineal figure A be given in species and magnitude, its
sides are given in magnitude.

Take a straight line BC given in position and magnitude, and
upon BC describe (18. 6.) the figure D similar, and similarly
placed, to the figure A, and let
EF be the side of the figure
A homologous to BC the side
of D ; therefore the figure D is
given in species. And because
upon the given straight line
BC, the figure D given in
species is described, D is given
(56. dat) in magnitude, and
the figure A is given in mag- »
nitude, therefore the ratio of
A to D is given : and the figure A is similar to D ; therefore the ratio
of the side EF to the homologous side BC is given (58. dat.) ; where-
fore (2. dat.) EF is given : and the ratio of EF to EG is given (3.
def ), therefore EG is given. And, in the same manner, each of the
other sides of a figure A can be shown to be given.

PROBLEM.

To describe a rectilineal figure A, similar to a given figure D, and
equal to another given figure H. It is prop. 25, b. 6, Elem.

Because each of the figures D, H is given, their ratio is given, which
may be found by making (cor. 45. 1.) upon the given straight line
BC the parallelogram BK equal to D, and upon its side CK making
(cor. 45. L) the parallelogram KL equal to H, and the angle ECL
equal to the angle MBC ; therefore the ratio of D to H, that is, of BK
to KL, is the same with the ratio of BC to CL : and because the figures
D, A are similar, and that the ratio of D to A, or H, is the same with the
ratio of BC to CL ; by the 58th dat. the ratio of the homologous sides
BC, EF is the same with the ratio of BC to the mean proportional
between BC and CL. Find EF the mean proportional ; then EF is the
side of the figure to be described, homologous to BC the side of D, and

Euclid's data. 331

the figure itself can be described by the 18th prop, book 6, which,
by the construction, is similar to D ; and because D is to A, as (2 cor.
20. 6.) BC to CL, that is, as the figure BK to KL ; and that D is equal
to BE, therefore A (14. 5.) is equal to KL, that is, to H.

PROP. LXI. 57.

Ir a parallelogram given in magnitude has one of its sides and
one of its angles given in magnitude, the other side also is given.*

Let the parallelogram ABDC given in magnitude, have the side
AB and the angle BAG given in magnitude, the other side AC is given.

Take a straight line EF given in position and magnitude ; and be-
cause the parallelogram AD is given in magni- A B
tude, a rectilineal figure equal to it can be
found (1. def.). And a parallelogram equal to

this figure can be applied (cor. 45. 1.) to the

given straight line EP in an angle equal to the q d

given angle BAG. Let this be the parallelo* E p

gram EFHG having the angle FEG equal to # j

the angle BAG. And because the parallelo- / /

grams AD, EH are equal and have the angles / /

at A and E equal ; the sides about them are ^ /

reciprocally proportioned (14. 6.); therefore q H

as AB to EF, so is EG to AG ; and AB, EF,
EG are given, therefore also AG is given (12. 6.). Whence the way
of finding AG is manifest.

PROP. LXII. H.

Ip a parallelogram has a given angle, the rectangle contained
by the sides about that angle has a given ratio to the parallelo-
gram.*

Let the parallelogram ABGD have the given
angle ABG, the rectangle AB, BG has a given
ratio to the parallelogram AG.

From the point A draw AE perpendicular to
BC ; because the angle ABG is given, as also the
angle AEB, the triangle ABE is given (42. dat.)
in species : therefore the ratio of BA to AE is
given. But as BA to AE, so is (1. 16.) the rect-
angle AB, BG to the rectangle AE, BG ; therefore
the ratio of the rectangle AB, BG to AE, BG, that
is, 35. 1.) to the parallelogram AG, is 'given.

And it is evident how the ratio of the rectangle to the parallelogram

* See Note.

t9% sccud's data.

nmy be fcmtA by making ttie angle FGH equal to tfee given an^
Afifc, and drawing from any point F In one of its sides, FK perpen-
dicular to the other GH : for GF is to FK, as BA to AE, that is, as
the rectangle AB, BC to the parallelogram AC.

C!oR. And if a triangle ABC has a given aiTgle ABC, the rectangle
AB, BC contained by the sides about that angle, shall have a given
ratio to the triangle ABC.

Complete the parallelogram. ABCD : therefore, by this proposition,
the rectangle AB, DC has a given ratio to the parallelogram AC ; and
AC has a given ratio to its half the triangle (41. 1.) ABC; therefore
the rectangle AB, BC has a given (9. dat.) ratio to the triangle ABC.

And the ratio of the rectangle to the triangle is found thus : make
the triangle FGK as was shown in the proposition ; the ratio of GF
to the half of the perpendicular FK is the same with the ratio of the
rectangle AB, BC to the triangle ABC. Because, as was shown, GF
is to FK, as AB, BC to the parallelogram AC ; and FK is to its hall,
as AC is to its half, which is the triangle ABC ; therefore, ex mqualh
GF is to the half of FK, as AB, BC rectangle is to the triangle ABC.

PROP. LXin. 66.

If two parallelograms be equiangular, as a side of the first to
a side of the second, so is the other side of the second to the
straight line to which the other side of the first has the same ratio
which the first parallelogram has to the second. And conse-
quently, if the ratio of the first parallelogram to the second be
given, the ratio of the other side of the first to that straight line
is given, and if the ratio of the other side of the first to that
straight line be given, the ratio of the first parallelogram to the
second is given.

Let AC, DF be two equiangular parallelograms ; as BC, a side of
the first, is to EF, a side of the second, so is DE the other side of the
second, to the straight line to which AB, the other side of the first,
has the same ratio which AC has to DF.

Produce the straight line AB. and make as BC to EF, so DE to
BG, and complete the parallelogram A

BGHC ; therefore because BC or GH is
to EF, as DE to BG, the sides about the
equal angles BGH, DEF are reciprocally
proportional ; wherefore (14. 6.) the pa-
rallelogram BH is equal to DF ; and AB
is to BG, as the parallelogram AC is to
BH, that is, to DF ; as therefore BC is
to EF, so is DE to BG, which is the
straight line to which AB has the same
ratio that AC has to DF.

.A-

* sugud's data. n3

And if the ratio of the paraTIelo^am AC to TO* be givBn, then the
ratio of the straight line AB to BG is given ; and if the ratio of AB
to the straight line BG be given, the ratio of the parallelogram AC to
DF is given.

PROP. LXIV. 74. 73.

Ir two paralle]<)grams have unequal but given angles, and if
as a side of the first to a side of the second, so the other side of
the second be made to a certain straight line ; if the ratio of the
first parallelogran) to the second be given, the ratio of the other
side of the first to that straight line shall be given. And if the
ratio of the other side of the first to that straight line be given,
the ratio of the first parallelogram to the second shall be
given.*

Let ABCD, EFGH be two parallelograms which have the unequal,
but given angles ABC, EFG ; and as BC to FG, so make EF to the
straight line M. If the ratio of the parallelogram AC to EG be given^
the ratio of AB to M is given.

At the point B of the straight line BC make the angle CBK equal
to the an^e EFG, and complete the parallelogram KBCL. And be*
cause the ratio of AC to EG is given, and that AC is equal (35. 1.)
to the parallelogram KC, therefore the ratio of KC to EG is given ;
and KC, EG are equiangular; therefore as BC to FG, so is (63. dat.)
EF to the straight line to which KB has a given ratio, viz. the same
•wrhich the parallelogram KC has to EG ; but as BC to FG, so is EF
to the straight line M ; therefore KB has a given ratio to M ; and
the ratio of AB to BK is given, because the triangle ABK is given
in species (43. dat.) ; therefore the ratio of AB to M is given (9.
dat.).

And if the ratio of AB to M be given, the ratio of the parallelo-
gram AC to EG is given; for since the ratio of KB to BA is
given, as also the ratio of AB to M, the ratio
of KB to M is given (9. dat.) ; and because
the parallelograms KC, EG are equiangular,
as BC to FG, so is (63. dat.) EF to the
straight line to which KB has the same ratio
which the parallelogram KC has to EG;
but as BC to FG, so is EF to M ; therefore
KB is to M, as the parallelogram KC is to
EG ; and the ratio of KB to M is given,
therefore the ratio of the parallelogram KC,
that is, of AC to EG, is given.

Cor. And if two triangles ABC, EFG, have two equal angles,
or two unequal, but given angles, ABC, EFG, and if as BC a side
of the first to FG a side of the second, so the other side of the se-

• See Note. i

334 suclid's data. *

cond EF be made to a straight line M ; if the ratio of the triangles
be given, the ratio of the other side of the first to the straight Jine
M is given.

Complete the parallelograms ABCD, EFGH ; and because the ratio
of the triangle ABC to the triangle EFG is given, the ratio of the
parallelogram AC to EG is given (15. 5.)* because the parallelograms
are double (41. 1.) of the triangles; and because EC is to FG, as
EF to M, the ratio of AB ^o M is given by the 63d dat. if the
angles ABC, EFG are equal ; but if they be unequal, but given
angles, the ratio of AB to M is given by this proposition.

And if the ratio of AB to M be given, the ratio of the parallelogram
AC to EG is given by the same proposition ; and therefore the ratio
of the triangle ABC to EFG is given.

PROP. LXV. 68.

If two equiangular parallelograms have a given ratio to one
another, and if one side have to one side a given ratio ; the other
side shall also have to the other side a given ratio.

Let the two equiangular parallelograms AB, CD have a given
ratio to one another, and let the, side EB have a given ratio to the
side FD ; the other side AE ^Jbas also a given ratio to the other
side CF.

Because^ the two equiangular parallelograms AB, CD have a
given ratio to one another ; as EB, a side of the first, is to FD, a
side of the second, so is (63. dat.) FC, the other side of the second,
to the straight line to which AE, the other side of the first, has
the same given ratio which the first parallelogram AB has to the
other CD. Let this straight line
be EG ; therefore the ratio of AE
to EG is given; and EB is to
FD, as FC to EG, therefore the
ratio of FC to EG is given, be-
cause the ratio of EB to FD is
given; and because the ratio of
AE to EG, as also the ratio of FC „ g-

to EG is given ; the ratio of AE
to CF is given (9. dat.).

The ratio of AE to CF may be found thus : take a straight line
H given in magnitude ; and because the ratio of the parallelogram
AB to CD is given, make the ratio of H to E the same with it. And
because the ratio of FD to EB is given, make the ratio of K to L the
same : the ratio of AE to CF is the same with the ratio of H to L.
Make as EB to FD, so FC to EG, therefore, by inversion, as FD to
EB, so is EG to FC ; and as AE to EG, so is (63. dat.) (the paral-
lelogram AB to CD, and so is) H to K ; but as EG to FC, so is (FD
to EB, and so is) K to L ; therefore ex sequalit as AE to FC, so is
H to L.

buclid's data. 336

PROP. LXVI. 69.

Ir two parallelograms have unequal, but given angles, and a
given ratio to one another ; if one side have to one side a given
ratio, the other side has also a given ratio to the other side.

Let the two parallelograms ABCD, EFGH which have the given
unequal angles ABC, EFG, have a given ratio to one another, and let
the ratio of BC to FG be given ; the ratio also of AB to EF is given.

At the point B of the straight line BC make the angle CBK equal
to the given angle EFG, and complete the parallelogram BKLC ; and
because each of the angles BAK, AKB, is given, the triangle ABK is
given (43. dat.) in species ; therefore the ratio of AB to BK is given;
and because, by the hypothesis, the ratio of the parallelogram AC to
EG is given, and that AC is equal (35. 1.) to BL; therefore the ratio
of BL to EG is given : and because BL is equiangular to EG, and, by
the hypothesis, the ratio of BC to FG is given ; therefore (65. dat.)
the ratio of KB to EF is given, and the A K D L

ratio of KB to BA is given ; the ratio there-
fore (9. dat.) of AB to EF is given.

The ratio of AB to EF may be found ^

thus : take the straight line MN given in E-.

position and magnitude ; and make the an- ^^
gle NMO equal to the given angle BAK,
and the angle MNO equal to the given an-
gle EFG or AKB: and because the paral- NO
lelogram BL is equiangular to FG, and has a given ratio to it,
and that the ratio of BC to FG is given ; find by the 65th dat. the
ratio of KB to EF ; and make the ratio of NO to OP the same with
it : tl^jBn the ratio of AB to EF is the same with the ratio of MO to
OP :'for since the triangle ABK is equiangular to MON, as AB to BK,
so is MO to ON : and as KB to EF, so is NO to OP ; therefore, ex
ssqualij as AB to EF, so is MO to OP.

PROP. LXVn. 70.

If the sides of two equiangular parallelograms have given
ratios to one another; the parallelograms shall have a given
ratio to one another.*

Let ABCD, EFGH be two equiangular parallelograms, and let the
ratio of AB to EF, as also the ratio of BC to FG, be given; the ratio
of the parallelogram AC to EG is given.

Take a straight line K given in magnitude, and because the

* Sec Note.

380 EUCUD*8 DATA.

ratio of AB to EF is given, make A D E H

the ratio of K to L the same with

it ; therefore L is given (2. dat.) :

and because the ratio of BC to

FG is given, make the ratio of L to

M the same : therefore M is given

(2. dat.) : and K is given, where-

ft)re (1. dat.) the ratio of K to M

is given : but the parallelogram AC is to the parallelogram EG, as

the straight line K to the straight line M, as is demonstrated in the

23d prop, of B. 6. Elem. ; therefore the ratio of AC to EG is given.

From this it is plain how the ratio of two equiangular parallrfo-
grams may be found when the ratios of then: sides are given.

PROP. LXVni. 70.

Ir the sides of two parallelograms which have unequal, but
given angles, have given ratios to one another; the parallelo-
grams shall have a given ratio to one another.*

Let two parallelograms ABCD, EFGH, which have the given un-
equal angles ABC, EFG, have the ratios of their sides, viz. of AB to
EF, and of BC to FG, given ; the ratio of the parallelogram AC to
EG is given.

At the point B of the straight line BC make the angle CBK equal
to the given angle EFG, and complete the parallelogram KBCL ;
and because each of the angles BAK, BEA is given, the triangle
ABK is given (43. dat.) in species : therefore the ratio of AB to BK
is given ; and the ratio of AB to EF is given ; wherefore (9. dat.)
the ratio of BK to EF is given : K A
and the ratio of BC to FG is
given ; and the angle KBC is equal
to the angle EFG ; therefore (67.
dat.) the ratio of the parallelo-
gram KC to EG is given: but
KC is equal (35. 1.) to AC ; there-
fore the ratio of AC to EG is
given.

The ratio of the parallelogram AC to EG may be found thus : take
the straight line MN given in position and magnitude, and make the
angle MNO equal to the given angle KAB, and the angle NMO equal
to the given angle AKB, or FEH : and because the ratio of AB to
EF is given, make the ratio NO to P the same ; also make the ratio
of P to Q the same with the given ratio of BC to FG, the parallelo-
gram AG is to EG, as MO to Q.

Because the angle KAB is equal to the angle MNO, and the
angle AKB equal to the angle NMO; the triangle AKB is equi-
angular to NMO : therefore as KB to BA, so is MO to ON ; and

* See Note.

rdclid'js data. S37

as BA to EF, so is NO to P ; wherdbre, ex xquedi, as KB to B^,
«o is MO to P : and BC is to F6» as P to Q, and the parallelograms
KC, EG are equiangular; therefore, as was shown in prop. 67, the
parallelogram KC, that is, AC is to EG, as MO to Gt.

Cor. 1. If two triangles, ABC, DEF have two equal angles, or
two unequal, but given angles, ABC, DEF, and if the ratios of the
sides about these angles, viz. the ratios of A G D H

AB to DE, and of BC to EF be given ; the
triangles shall have a given ratio to one
another.

Complete the parallelograms BG, EH;
the ratio of BG to. EH is given (67. or 68.
dat); and therefore the triangles which are the halves (34. 1.) of
them have a given (15. 6. 72.) ratio to one another.

CoR. 2. If the bases BC, EF of two triangles ABC, DEF have a
given ratio to one another, and if also the straight lines AG, DH
which are drawn to the bases from the opposite angles, either in
equal angles, or unequal, but given angles, AGC, DHF have a given
ratio to one another ; the triangles c- ^
shall have a given ratio to one an-
other.

Draw BK, EL parallel to AG,
DH, and complete the parallelo-
grams KC« LF. And because the
angles AGC, DHF, or their equals, the angles KBC, LEIF are either
equal, or unequal, but given ; and that the ratio of AG to DH, that
is, of KB to LE, is given, as also the ratio of BC to EF ; therefore
(67. or 68. dat.) the ratio of the parallelogram KC to LF is given ;
wherefore also the ratio of the triangle ABC to DEF is given
(41. 1. 15. 5.)

PROP. LXIX. 61.

If a parallelogram which has a given angle be applied to one
side of a rectilineal figure given in species; if the figure have a
given ratio to the parallelogram, the parallelogram is given in
species.

Let ABCD be a rectilineal figure given in species, and to one side
of it AB, let the parallelogram ABEF, having the given angle ABE,
be applied ; if the figure ABCD have a given ratio to the parallelo-
gram BF, the parallelogram BF is given in species.

Through the point A draw AG parallel to BC, and through the
point C draw CG parallel to AB, and produce GA, CB to the points H,
K: because the angle ABC is given (3. def.), and the ratio of AB to BC
is given, the figure ABCD being given in species ; therefore the pa-
rallelogram BG is given (3. def.), in species. And because upon the
same straight line AB the two rectilineal figures BD, BG given in spe-
cies are described, the ratio of BD to BG is given (53. dat.) ; and, by
hypothesis, the ratio of BD to the parallelogram BFis given ; wherefore
(9. dat.) the ratio of BF, that is, (35. 1 .) of the parallelogram BH, to BG is

388

BUCLID 8 DATA.

given, and therefore (1. 6.) the ratio of the straight line KB to BC is
given ; and the ratio of BC to BA is given, wherefore the ratio of
KB to BA is given (9. dat.) : and because the angle ABC is given,
the adjacent angle ABK is given; and the angle ABE is given,
therefore the remaining angle KBE is given. The angle EKB is
also given, because it is equal to the angle ABK; therefore the tri-
angle BKE is given in species, and consequently the ^atio of EB to
BK is given ; and the ratio of KB to BA is given, wherefore (9. dat.)
the ratio of EB to BA is given ;
and the angle ABE is given,
therefore the parallelogram BP
is given in species.

A parallelogram similar to
BF may be found thus : take a
straight line LM given in posi-
tion and magnitude: and be-
cause the angles ABK, ABE
are given, make the angle NLM equal to ABK, and the angle NLO
equal to ABE. And because the ratio of BF to BD is given, make
the ratio of LM to P the same with it ; and because the ratio of the
figure BD to BG is given, find this ratio by the 5Bd dat. and make
the ratio of P to Q, the same. Also, because the ratio of CB to BA
is given, make the ratio of Q, to R the same ; and take LN equal to
R ; through the point M drawn OM parallel to LN, and complete the
parallelogram NLOS; then this is similar to the parallelogram BF.

Because the angle ABK is equal to NLM, and the angle ABE
to NLO, the angle KBE is equal to MLO ; and the angles BKE,
LMO are equal, because the angle ABK is equal to NLM ; therefore
the triangles BKE, LMO are equiangular to one another ; wherefore
as BE to BK, so is LO to LM : and because as the figure BF to BD,
so is the straight line LM to P ; and as BD to BG, so is P to Q, ;
ex sequalh as BF, that is (35. L) BH to BG, so is LM to Q, : but BH
is to (1. 6.) BG, as KB to BC : as therefore KB to BC, so is LM to
*Cl ; and because BE is to BK, as LO to LM ; and as BK to BC, so
is LM to d : and as BC to BAi so Q, was made to R ; therefore ex
mqucUi, as BE to BA, so is LO to R, that is to LN ; and the angles
ABE, NLO are equal ; therefore the parallelogram BF is similar
toLS.

PROP. LXX.

62.78.

If two straight lines have a given ratio to one another, and
upon one of them be described a rectilineal figure given in
species, and upon the other a parallelogram having a given
angle ; if the figure have a given ratio to the parallelogram, the
parallelogram is given in species.*

Let the two straight lines AB, CD have a given ratio to one
another, and upon AB let the figure AEB given in species be de-

* See Note.

Euclid's data. , ?W

scribed, and u^)n CD the parallelogram DP having the given angle
FCD ; if the ratio of AEB to DF be given, the parallelogram DF is
given in species.

Upon the straight line AB, conceive the parallelogram AG to be
described similar, and similarly placed to FD; and because the ratio
of AB to CD is given, and upon them are described the similar rec-
tilineal figures* AG, FD; the ratio of
AG to FD is given (54. dat.) ; and the
ratio of FD to AEB is given; there-
fore (9. dat.) the ratio of AEB to
AG is given ; and the angle ABG is
given, because it is equal to the angle
FCD : because therefore the parallelo-
gram AG which has a given angle
ABG is applied to a side AB of the
figure AEB given in species, and the

ratio of AEB to AG is given, the parallelogram AG is given (69. dat.)
in species; but FG is similar to AG; therefore FD is given in
species.

A parallelogram similar to FD may be found thus : take a straight
line H given in magnitude ; and because the ratio of the figure AEB
to FD is given, make the ratio of H to K the same with it : also,
because the ratio of the straight line CD to AB is given, find by the
64th dat. the ratio which the figure FD described upon CD has to
the figure AG described upon AB similar to FD ; and make the ratio
of K to L the same with this ratio ; and because the ratios of H to K,
and of K to L are given, the ratio of H to L is given (9. dat) ;
because, therefore, as AEB to FD, so is H to K ; and as FG to AG,
so is K to L ; ex sequalU as AEB to AG, so is H to L ; therefore the
ratio of AEB to AG is given ; and the figure AEB is given in species,
and to its side AB the parallelogram AG is applied in the given
angle ABG ; therefore by the 69th dat. a parallelogram may be found
similar to AG : let this be the parallelogram MN ; MN also is similar
to FD ; for, by the construction, MN is similar to AG, and AG is
similar to FD ; therefore the parallelogram FD is similar to MN.

PROP. LXXI. 81.

If the extremes of three proportional straight lines have given
ratios to the extremes of other three proportional straight lines ;
the means shall also have a given ratio to one another: and if one
extreme have a given ratio to one extreme, and the mean to the
mean ; likewise the other extreme shall have to the other a given
ratio.

Let A, B, C be three proportional straight lines, and D, E, F three
other ; and let the ratios of A to D, and of C to F be given ; then the
ratio of B to E is also given.

Because the ratio of A to D, as also of C to F is given, the ratio

i*&

BVCKID'S DATA.

I
C

of the rectangle A, C to the rectangle D, P is ^Teii^67. dat); but
the square of B is equal (17. 6.) to the rectangle A, C ; and the sqi»re
of E to the rectangle (17. 6.) D, F ; therefore the ratio of the square
of B to the square of E is given ; wherefore (5H dat.) also the ratio of
the straight line B to E is given.

•Next, let the ratio of A to D,- and of B to
E be given; then the ratio of C to F is also
given.

Because the ratio of B to E is given, the ratio
of the square of B to the square of E is given
(54 dat.); therefore (17. 6.) the ratio of the rec-
tangle A, C to the rectangle D, F is given ; and
the ratio of the side A to the side D is given ;
therefore the ratio of the other side C to the other
F is given (65. dat).

Cor. And if the extremes of four proportionals have to the ex-
tremes of four other proportionals given ratios, and one of the means
a given ratio to one of the means ; the other mean shall have a given
ratio to the other mean, as may be shown in the same manner as
in the foregoing proposition.

E F

PROP. LXXII.

82.

Ir four straight lines be proportionals ; as the first is to the
Qtraight line to which the second has a given ratio, so is the
third to a straight line to which the fourth has a given ratio.

Let A, B, C, D be proportional straight lines, viz. as A to B, so C
to D ; as A is to the straight line to which B has a given ratio so is
C to a straight line to which D has a given ratio.

Let E be the straight line to which B has a given
ratio, and as B to E, so make D to F : the ratio of
B to E is given (Hyp.), and therefore the ratio of D
to F ; and because as A to B, so is C to D ; and as
B to E, so D to F ; therefore, ex sequalh as A to E,
so is C to F ; and E is the straight line to which B
has a given ratio, and F that to which D has a given
ratio ; therefore as A is to the straight line to which
B has a given ratio, so is C to a line to which D has
a given ratio.

A B
C D

E
F

PROP. Lxxni.

88.

If four straight lines be proportionals; as the first is to
the straight line to which the second has a given ratio, so
is a straight line to which the third has a given ratio to the
fourth.*

♦ See Note.

EUCLID'8 DATA.

B
C

E
D

64.

Let the straight line A be to B, aa C to D: as A
to the straight line to which B has a given ratio, so
Is a straight line to which C has a given ratio to D»

Let E be the straight line to which B has a given
ratio, and as B to E, so make F to C ; because the
ratio of B to E is given, the ratio of C to F is given :
and because A is to B, as C to D ; and as B to E,
so F to C ; therefore, ex sequali in proportione per-
SurbtiUi (23. 5.), A is to E, as F to D ; that is, A is
to E, to which B has a given ratio, as F, to which
C has a given ratio, is to D.

PROP. LXXIV.

Ir a triangle have a given obtuse angle ; the excess of the
square of the side which subtends the obtuse angle, above the
squares of the sides which contain it, shall have a given ratio to
the triangle. «

Let the triangle ABC have a given obtuse angle ABC ; and pro-
dace the straight line CB, and from the point A draw AD perpendi-
cular to BC : the excess of the square of AC above the squares of
AB, BC, that is (12. 2.), the double of the rectangle contahaed by
DB, BC, has a given ratio to the triangle ABC.

Because the angle ABC is given, the angle ABD is also given ;
and the angle ADB is given ; wherefore the triangle ABD is given
(43. dat.) in species ; and therefore the ratio of AD to DB is given :
and as AD to DB, so is (1. 6.) the rectangle AD, BC to the rectangle
DB, BC ; wherefore the ratio of the rectangle AD, BC to the rect-
angle DB, BC is given, as also the ratio of twice the rectangle DB,
BC to the rectangle AD, BC : but the ratio of A E

the rectangle AD, BC to the triangle ABC is
given, because it is double (41. L) of the tri-
angle ; therefore the ratio of twice the rectan-
gle DB, BC to the triangle ABC is given (9.
dat.) ; and twice the rectangle DB, BC is the
excess (12. 2.) of the square of AC above the D B C

squares of AB, BC ; therefore this excess has a given ratio to the
triangle ABC.

And the ratio of this excess to the triangle ABC may be found
thus : take a straight line EF given in position and magnitude ; and
because the angle ABC is given, at the point F of the straight line
EF, make the angle EFG equal to the angle ABC ; produce GF, and
draw EH perpendicular to FG ; then the ratio of the excess of the
square of AC above the squares of AB, BC to the triangle ABC, is
the same with the ratio of quadruple the straight line HF to HE.

Because the angle ABD is equal to the angle EFH, and the
angle ADB to EHF, each being a right angle : the triangle ADB
is equiangular to EHF ; therefore (4. 6.) as BD to DA, so FH to
HE; and as quadruple of BD to DA, so is (cor. 4. 5.) quadruple
of FH to HE: but as twice BD is to DA, so is (1. 6) twice the
rectangle DB, BC to the rectangle AD, BC; and as DA to the

84t , buvud's data^

half of it, so is (cor. 5.) the rectangle AB, BC to its half the triangle
ABC ; therefore, ex mqualh as twice BD is to the half of DA, that is,
as quadruple of BD is to DA, that is, as quadruple of FH*to HE, so
is twice the rectangle DB, BC to the triangle ABC.

PROP. LXXV. 65.

Ir a triangle have a given acute angle, the space by which
the square of the side subtending the acute angle is less than the
squares of the sides which contain it, shall have a given ratio to
the triangle.

Let the triangle ABC have a given acute angle ABC, and draw
AD perpendicular to BC ; the space by which the square of AC is
less than the squares of AB, BC, that is (13. 2.), the double of the
rectangle contained by CB, BD, has a given ratio to the triangle
ABC. •

Because the angles ABD, ADB are each of them given, the tri-
angle ABD is given in species ; and therefore the ratio of BD to
DA is given : and as BD to DA, so is the rect- A

angle CB, DB, to the rectangle CB, AD ; there-
fore the ratio of these rectangles is given, as
also the ratio of twice the rectangle CB, BD to
the rectangle CB, AD ; but the rectangle CB,
AD has a given ratio to its half the triangle
ABC ; therefore (9. dat.) the ratio of twice the B DC

rectangle CB, BD to the triangle ABC is given ; and twice the rect-
angle CB, BD is (13. 2.) the space by which the square of AC is
less than the squares of AB, BC ; therefore the ratio of this space to
the triangle ABC is given : and the ratio may be found as in the
preceding proposition.

LEMMA.

If from the vertex A of an isosceles triangle ABC any straight
line AD be drawn to the base BC, the square of the side AB is equal
to the rectangle BD, DC of the segments of the base together with
the square of AD; but if AD l^e drawn to the base produced, the
square of AD is equal to the rectangle BD, DC together with the
square of AB.

Case 1. Bisect the base BC in E, and join
AE which will be perpendicular (8. 1.) to
BC ; wherefore the square of AB is equal (47.
1.) to the squares of AE, EB; but the square
of EB is equal (5. 1.) to the rectangle BD,
DC, together with the square of DE; there- p n T?

fore the square of A B is equal to the squares ^ But*
of AE, ED, that is to (47. 1.) the squares of AD, together with the
rectangle BD, DC ; the other case is shown in the same way by 6.
2. Elem.

buclid'8 data.

•48

PROP. LXXVL 67.

Ir a triangle have a given angle, the excess of the square of
the straight line which is equal to the two sides that contain the
given angle, above the square of the third side, shall have a given
ratio to the triangle.

Let the triangle ABC have the given angle BAG ; the excess of
the square jf the straight line which is equal to BA, AC together
above the square of BC, shall have a given ratio to the triangle ABC.

Produce BA, and take AD equal to AC ; join DC, and produce
it to E, and through the point B draw BE parallel to AC; join
AE, and draw AF perpendicular to DC ; and because AD is equal
to AC, BD is equal to BE: and BC is drawn from the vertex B
of the isosceles triangle DBE ; therefore, by the lemma, the square
of BD, that is, of BA and AC together, is equal to the rectangle
DC, CE together with the square of BC ; and therefore, the square
of BA, AC, together, that is, of BD : is
greater than the square of BC by the
rectangle DC, CE; and this rectangle
has a given ratio to the triangle ABC :
because the angle BAC is given, the
adjacent angle CAD is given : and each
of the angles ADC, DCA is given, for
each of them is the half (5. & 32.) of
the given angle BAC ; therefore the tri-
angle ADC is given (43. dat.) in spe-
cies; and AP is drawn from its ver-
tex to the base in a given angle ; wherefore the ratio of AP to the
base CD is given (50. dat.)*; and as CD to AF, so is (1. 6.) the
rectangle DC, CE to the rectangle AF, CE ; and the ratio of the
rectangle AP, CE to its half (41. 1.), the triangle ACE, is given;
therefore the ratio of the rectangle DC, CE to the triangle ACE, that
is (37. 1.), to the triangle ABC, is given (9. dat.), and the rectangle
DC, CE is the excess of the square of BA, AC together above the
square of BC : therefore the ratio of this excess to the triangle ABC
is given. •

The ratio which the rectangle DC, CE has to the triangle ABC
is found thus: take the straight line GH given in position and
magnitude, and at the point G in GH make the angle HGK equal
to the given angle CAD, and take GK equal to GH ; join KH, and
draw GL perpendicular to it : then the ratio of HK to the half of
GL is the same with the ratio of the rectangle DC, CE to the tri-
angle ABC : because the angles HGK, DAC, at the vertices of the
isosceles triangles GHK, ADC are equal to one another, these tri-
angles are similar ; and because GL, AF are perpendicular to the
bases HK, DC, as HK to GL, so is (4. 6. 22. 5.) (DC to AF, and
so is) the rectangle DC, CE to the rectangle AF, CE ; but as GL
to its half, so is the rectangle AF, CE to its half, which is the tri-
angle ACE, or the triangle ABC; therefore, ex ssgvali, HK is to
the half of the straight line GL, as the rectangle DC, CE is to the
triangle ABC.

844

BOCUD*a DATA.

Cor. And if a triangle have a given angle, the space by which
the square of the straight tine which is the difference of the sides
which contain the given angle is less than the square of the third
side, shall have a given ratio to the triangle. This is demonstrated
the same way as the preceding proposition, by help of the second
case of the lemma.

PROP. LXXVII. ^ L.

Ip the perpendicular drawn from a given angle of a triaogle
to the opposite side, or base, have a given ratio to the base, the
triangle is given in species.*

Let the triangle ABC have the given angle BAC, and let the per-
pendicular AD drawn to the base BC have a given ratio to it: the
triangle ABC Is given in species.

If ABC be an isosceles triangle, it is evident (5. and 32. 1.) that

R D C

if any one of its angles be given, the rest are also given ; and there-
fore the triangle is given in species, without the consideration of the
ratio of the perpendicular to the base, which in this case is given by
prop. 50.

But when AEtt! is not an isosceles triangle, take any straight
line £F given in position and magnitude, and upon it describe
the segment of a circle EGF containing an angle equal to the
given angle BAC; draw GH bisecting EF at right angles, and
join EG, GF: then, since the angle EGF is equal to the angle
BAC, and that EGF is an isosceles triangle, and ABC is not, the
angle FEG is not equal to the angle CBA : draw EL, making ^e
»ngle FEL equal, to the angle CBA ; join FL, and draw LM per-
pendicular to EF; then, because the triangles ELF, BAC are
equiangular, as also are the triai^les MLE, DAB, as ML to LE,
so is DA to AB ; and as LE to EF, so is AB to BC ; wherelbre,
€x «guaii, as LM to EF, so is AD to BC ; and because the ratio
of AD to BC is given, therefore the ratio of LM to EF is given;
and EF is given, wherefore (2. dat) LM also is given. Complete
the paralleiogram LMFK ; and because LM is given, FK is given
in magnitude ; it is also given in position, and the point F is
given, and consequently (30. dat.) the point K; and because
through K the straight line KL is drawn parallel to EF which is
gilven in position, therefore (31. dat.) KL is given in position:
and the circumference ELF is given in position; tliereibre the
point L is given (28. dat.). And because the pdnts L, E, F, are

* See Note.

■volid's data.

949

given, the etraight lines LE, EF, FL, are given (20. dat) in mag-
nitude ; therefore the triangle LEF is given in species (42. dat) ;

E O H M P

and the triangle ABC is similar to LEF, wherefore also ABC is given
in species.

Because LM is less than GH, the ratio of LM to EF, that is, the
given ratio of AD to BC, must be less than the ratio of GH to EF,
-which the straight line, in a segment of a circle containing an angle
equal to the given angle, that bisects the base of the segment at
right angles, has unto the base.

Cor. 1. If two triangles, ABC, LEF, have one angle BAC equal
to one angle ELF, and if the perpendicular AD be to the base BC,
as the perpendicular L^ to the base EF, the triangles ABC, LEF
are similar.

Describe the circle EFG about the triangle ELF, and draw LN
parallel to EF; join EN, NF, and draw NO perpendicular to EF;
because the angles ENF, ELF are equal ; and that the angle EFN
is equal to the alternate angle FNL, that is, to the angle FEL in the
same segment ; therefore the triangle NEF is similar to LEF ; and
in the segment EGF there can be no other triangle upon the base
EF, which has the ratio .of its perpendicular to that base the same
with the ratio of LM or NO to EF, because the perpendicular must
be greater or less than LM or NO; but, as has been shown in the
preceding demonstration, a triangle similar to ABC can be described
in the segment EGF upon the base EF, and the ratio of its perpen-
dicular to the base is the same, as was there shown, with the ratio
of AD to BC, that is, of LM to EF ; therefore that triangle must be
either LEF, or NEF, which therefore are similar to the triangle ABC.

CoR. 2. If a triangle ABC have a given angle BAC, and if the
straight line AR drawn from the given angle to the opposite side BC,
in a given angle ARC, have a given ratio to BC, the triangle ABC is
given in species.

Draw AD perpendicular to BC ; therefore the triangle ARD is
given in species ; wherefore the ratio of BD to AR is given ; and the
ratio of AR to BC is given, and consequently (9. dat.) the ratio of
AD to BC is given ; and tbe triangle ABC is therefore given in
species (77. dat.)

CoR. 3. If two triangles ABC, LEF have one angle BAC equal
to one angle ELF, and if straight lines drawn from these angles
to the bases, making with them given and equal angles, have the
same ratio to the bases, each to each ; then the triangles are simi-
lar ; lor having drawn perpendiculars to the bases from the equal
angles, as one perpendicular is to its base, so is the other to its
base (4. 6. 22. 6.); wherefore, by Cor. 1. the triangles are similar.

9^9

IDGUD*8 PATA*

A triangle sknOar to AfiC may be found thus : liaving dfisciibed
the segment EOF, and drawn the straight line GH, as was directed
in the proposition, find FK, which has to EF the given ratio of AD to
BC ; and place FK at right angles to EF from the point F ; then,
because, as has been shown, the ratio of AD to BC, that is of FK to
EF, must be less than the ratio of GH to EF ; therefore FK is less
than GH ; and consequently the parallel to EF, drawn through the
point K, must meet the circumference of the segment in two points :
let L be either of them, and join EL, LF, and draw LM perpendicu-
lar to EF : then, because the angle BAG is equal to the angle ELF,
and that AD is to BC, as KF, that is. LM to EF, the triangle ABC
is similar to the triangle LEF, by Cor. 1.

PROP. Lxxvm.

80.

If a triangle have one angle given» and if tlie ratio of the rect-
angle of the sides ^hich contain the given angle to the square of
the third side be given, the triangle is given in species*

Let the triangle ABC have the given angle BAC, and let the ratio
of the rectangle BA, AC to the square of BC be given ; the triangle
ABC is given in species.

From the point A, draw AD perpendicular to BC ; the rectangle
AD, BC has a given ratio to its half (41. l.)> the triangle ABC; and
because the angle BAC is given, the ratio of the triangle ABC to the
rectangle BA, BC is given (Cor. 62. dat.).; and by the hypothesis,
the ratio of the rectangle BA, AC to the square of BC is given ;
therefore (0. dat.) the ratio of the rectangle AD, BC to the square of
BC, that is (1, 6.), the ratio of the straight line AD to BC is given ;
wherefore the triangle ABC is given in species (77. dat.).

A triangle similar to ABC may be found thus : take a straight line
EF given in position and magnitude, and make the angle FEG equal
to the given angle BAC, and draw FH perpendicular to EG, and BK
perpendicular to AC : therefore the triangle ABK, EFH are similar,
and the rectangle AD, BC, ^ M
or the rectangle BK, AC
which is equal to it, is to

B D N

the rectangle BA, AC, as
the straight line BK to BA,
that is, as FH to FK Let
the given ratio of the rect-
angle BA, AC to the square
of BC be the same with the ratio of the straight line EF to FL ; there-
fore, ex sequali, the ratio of the rectangle AD, BC to the square of BC,
that is, the ratio of the straight line AD to BC, is the same with the
ratio of HF to FL ; and because AD is not greater than the straight
line MN in the segment of the circle described about the triangle ABC,
which iHsects BC at right angles ; the ratio of AD to BC. that is, of HF

EIICU&*8 DATA. M7

ta FL, must not be grei^r than the ratio of MN ta BC : let it be
so; and, by the 77th dat. find a triangle OPQ, which has one of
its angles POQ, equal to the given angle BAG, and the ratio of the
perpendicular OR, drawn from that angle to the base Pd the same
with the ratio of HF to FL ; then the triangle ABC is similar to
OPQ, : because, as has been shown, the ratio of AD to BC is the
same with the ratio of (HF to FL, that is^ by the construction, with
the ratio of) OR to PQ; and the angle BAG is equal to the ailgle
FOCI; therefore the triangle ABC is similar (1. Cor. 77. dat.) to the
triangle OPQ*

Otherwise^

Let the triangle ABC have the given angle BAG, and let the ratio
of the rectangle BA, AC, to the square of BC be given; the triangle
ABC is given in species.

Because the angle BAG is given, the excess of the square of
both the sides BA, AC together above the square of the third
side BC has a given (76. dat.) ratio to the triangle ABC. Let the
figure D be equal to this excess ; therefore the ratio of D to the
triangle ABC is given : and the ratio of the triangle ABC to the
rectangle BA, AG is given (Cor. 62. dat.)» because BAG is a given
angle; and the rectangle BA, AC has a
given ratio to the square of BC : where- Jn^

fore (10. dat.) the ratio of D to the square
of BC is given; and by composition (7. dat.)
the ratio of the space D together with the
square of BO to the square of BC is given ; B G

but D together with the square of BC is

equal to the square of both BA and AC together ; therrfore the ratio
of the square of BA, AG together to the square of BC is given ; and
the ratio of BA, AC together to BC is therefore given (59* dat.) ; and
file angle BAG is given, wherefore (48. dat.) the triangle ABC is
given in species.

The composition of this, which depends upon those of the 76th
and 48th propositions, is more complex than the preceding composi-
tion, which depends upon that of prop. 77, which is easy.

PROP. LXXIX. K.

Ir a triangle have a given angle, and if the straight line
drawn from that angle to the base, making a given angle with
it, divide the base into segments which have a given ratio to one
another ; the triangle is given in species.*

Let the triangle ABC have the given angle BAG, and let the
straight line AD drawn to the base BC making the given angle
ADB, divide BC into the segments BD, DC which have a given ratio
to one another ; the triangle ABC is given in species.

Describe (6. 4.) the circle BAG about the triangle, and frotti its

* Sec Note.

-TV

348 KDCLIo'g DATA.

centre E, draw BA, SB, EC, ED; because the angle BAG is
fiven, the angle BEG at the centre, which is the double (20. 3.)
of it, is given. And the ratio of BE to EC is given, because thejr
are equal to one another ; therefore (44. dat.) the triangle EEC is
given in species, and the ratio of EB to BC is given; aiso the
ratio of CD to BD is given (7. dat.), because the ratio of BD to
DC is given; therefore the ratio of EB to BD is given (9. dat.)
and the angle EBC is given, wherefore the triangle EBD is given
(9. dat.) in species, and the ratio of EB, that is, of EA to ED, is
therefore given; and the angle EDA is given, because each of the
angles BDE, BDA is given ; therefore the triangle AED is given
(47. dat.) in species, and the angle AED given :
also the angle DEC is given, because each of
the angles BED, BEC is given; therefore the
angle AEG^is given, and the ratio of EA to
EC, which are equal, is given ; and the trian-
gle AEC is therefore given (44. dat.) in species, B
and the angle EGA given; and the angle EGB .
is given, wherefore the angle ACB is given, and
the angle BAG is also given; therefore (43. dat.) the triangle ABC is.
given in species.

A triangle similar to ABC may be found, by taking a straight line
given in position and magnitude, and dividing it in the given ratio
which the segments BD, DC are required to have to one another :
then, if upon that straight line a segment of a circle be described
containing an angle equal to the given angle BAG, and a straight
line be drawn from the point of division in an angle equal to the
given angle ADB, and from the point where it meets the circum*
ference, straight lines be drawn to the extremity of the first line,
these, together with the first line, shall contain a triangle similar to
ABC, as may easily be shown.

The demonstration may be also made in the manner of that of the
77th prop, and that of the 77th may be made in the manner of this.

PROP. LXXX. L.

If the sides about an angle of a triangle have a given ratio
to one another, and if the perpendicular drawn from that angle
to the base have a given ratio to the base; the triangle is given
in species.

Let the sides BA, AG, about the angle BAG of the triangle ABC
have a given ratio to one another, and let the perpendicular AD have
a given ratio to the base BC ; the triangle ABC is given in species.

First, let the sides AB, AC be equal to one another, therefore
the perpendicular AD bisects (26. 1.) the base 1 A
BC ; and the ratio of AD to BC, and therefore to
its half DB, is given; and the angle ADB is given;
wlierefore the triangle (43. dat.) ABD, and conse-
quently the triangle ABC, Is given (44. dat.) in
species.

EUCLID*! DATA. 349

Bat let the sides be unequal, and BA be greater th^n AG ; and
make the angle GAE, equal to the angle AEG ; because the angle
AEB is common to the triangles AEB, GEA, they are similar;
therefore as AB to BEl, so is GA to AE, and, by permutation, as
BA to AO^ so is BE to EA, and so is EA to EG ; and the ratio of
BA to AG is given, therefore the ratio of BE to EA, and the ratio
of EA to EG, as also the ratio of BE to EG is given (9. dat.);
wherefore the ratio of EB to BG is given (6. dat.) ; and the ratio
of AD to BG is given by the hypothesis, there-
fore (9. dat.) the ratio of AD to BE is given ;
and the ratio of BE to EA was shown to be
given \ wherefore the ratio of AD to EA is given,
and ADE is a right angle, therefore the triangle
ADE is given (46. dat.) in species, and the angle
AEB given; the ratio of BE to EA is likewise
given, therefore (44. dat.) the triangle ABE is given in species, and
consequently the angle EAB, as also the angle ABE, that is, the
angle GAE, is given; therefore the angle BAG is given, and the
angle ABG being also given, the triangle ABG is given (43. dat.)
in species.

How to find a triangle which shall have the things which are
mentioned to be given in the proposition, is evident in the first
^ase; and to find it more easily in the other case, it is to be
observed, that, if the straight line EF equal to EA be placed in
£B towards B, the point F divides the base BG into the segments
BF, FG which have to one another the ratio of the sides BA, AG»
because BE, EA, or EF, and EG, were shown to be proportionals^
therefore (19. 5.) BF is to FG as BE to EF, or EA, that is, as BA
to AG ; and AE cannot be less than the altitude of the triangle
ABGy "but it may be equal to it, which, if it be, the triangle, in this
ease, as also the ratio of the sides, may be thus found: having
given the ratio of the perpendicular to the base, take the straight
fine GH given in position and magnitude, for the base of the tri-
angle to be found ; and let the given ratio of the perpendicular to
the base be that of the straight line K to GH, that is, let R be
equal to the perpendicular; and suppose GLH to be the triangle
which is to be found, therefore having made the angle HLM equal
to LGH, it is required that LM be perpendicular to GM, and
equal to E ; and because GM, ML, MH are proportionals, as was
shown of BE, EA, EG, the rectangle GMH is equal to the square
of ML. Add the common square of NH, (having bisected GH in N),
and the square of NM is equal (6. 2.) to the squares of the
given straight lines NH and ML or K ; therefore the square of NM
and its side NM, is given, as also the point M, viz. by taking the
straight line NM, the square of which is equal to the squares of
NH, ML. Draw ML equal to K, at right angles to GM ; and be-
cause ML is given in position and magnitude, therefore the point L
is given; join LG, LH; then the triangle LGH is that which was to
be found, for the square of NM is equal to the squares of NH and
ML, and taking away the common square of NH, the rectangle GMH

360

EUCLID'8 DATA.

NCIH

iB equal (6. 8.) to the square of
ML, therefore as GM to ML,
80 is ML to MH, and the tri-
angle LGM is (6. 6.) therefore
equiangular to HLM, and the
angle tiLM equal to the an-
gle GLM, and the straight line
LM drawn from the vertex of
the triangle making the angle
HLM equal to LGH, is perpen-
dicular to the base and equal to the given straight line K, as was
required ; and the ratio of the sides GL, LH is the same with the
ratio of GM to ML, that is, with the ratio of the straight line which
is made up of GN the half of the given base and of NM, the square
of which is equal to the squares of GN and K, to the straight line K.

And whether this ratio of GM to ML be greater or less than the
ratio of the sides of any other angle upon the base GH, smd of which
the altitude is equal to the straight line K, that is, the vertex of
which is in the parallel to GH drawn through the point L« may be
thus found. Let OGH be any such triangle, and draw 0P| milking
the angle HOP equal to the angle OGH ; therefore, as before, GP,
PO, PH are proportionals, and PO cannot be equal to LM, because
the rectangle GPH would be equal to the rectangle GMH, which is
impossible ; for the point P cannot fall upon M, because O would
then ^11 on L; nor can PO be less than LM, therefore it is greater;
and consequently the rectangle GPH is greater than the rectan^e
QMH, and the straight Jine GP greater than GM: therefore, the
ratio of GM to MH is greater than the ratio of GP to PH, aad
the ratio of the square of GM to the square of ML is therdbre
(2. cor. 20. 6.) greater than the ratio of the square of GP to
the square of PO, and the ratio of the straight line GM to ML,
greater than tk^ ratio of GP to PO. But as GM to ML, so is.GL
to LH ; and as GP to PO, so is GO to OH ; therefore the ratio of
GL to LH is greater than the ratio of GO to OH ; wherefore the
ratio of GL to LH is the greatest of all others ; and consequently
the given ratio of the greater side to the less, must not be greater
than this ratio.

But if the ratio of the sides be not the same with this greatest
ratio of GM to ML, it must necessarily be less than it : let any less
ratio be given, and the same things being supposed, viz. that GH
is the base, and K equal to the altitude of the triangle, it may be
found as follows. Divide GH in the point Q, so that the ratio of
GO, to QH may be the same with the given ratio of the sides;
and as GQ, to QH, so make GP to PO, and so will (19. 5.) PQ, be
to PH; wherefore the square of GP is to the square of PQ^ as
(2. cor. 29. 6.) the straight line GP to PH : and because GM, MU
MH are proportionals, the square of GM is to the square of ML,
as (2. cor. 20. 6.) the straight line GM to MH: but the ratio of
GQ, to QH, that is, the ratio of GP to PO, is less than the ratio of
GM to ML ; and therefore the ratio of the square of GP to the
square of PQ is less than the ratio of the square of GM to that of

isuciiii)*a D4ti. aftl

ML; 9nd oontaquently the ratio of the straight line OP to PH Is
les^ than the ratio of GM to MH; and by division, the ratio of
GH to HP is less than that of GH to HM ; wherefore (10. 6.) the
straight line HP is greater than HM« and the rectangle GPH, that
i9> the square of PQ, greater than the rectangle GMH, that is, than
the square of ML, and the straight line P€l is therefeore greater
than ML. Draw LH parallel to GP, and from P draw PR at
right angles to GP : because PQ, la greater than ML or PR, the
circle described from the centre P, at the distance PQ, must ne-
cessarily cut LR in two points ; let these be Q, 8 and join OQ,
OH ; SG, SH : each of the triangles OGH» SGH has the things
mentioned to be given in the proposition: join OP, SP; and be*
ca^se aa GP to BQ, or PQ, ao is PC to PH, the triangle OGP ts
equiangular to HOP ; as, therefore, OG to GP, so is HO to OP ;
and, by permutation, aa GO to OH, so is GP to PO, or PQ; and
so is GO, toQH: therefore the triangle OGH has the ratio c^ its
sides GO, OH the same with the given ratio of GQ to QH : and the
perpendicular has to the base the given ratio of E to GH, because
the perpendicular is equal to LM or K : the like may be shown in '
the same way of the triangle SGH«

This construction, by which the triangle OGH is found is shorter
than that which would be deduced from the demonstration of the
datum, by reason that the base GH is given in position and magni*
tude, which was not supposed in the demonstration; the same thing
is to l)e observed in the next proposition.

PROP. LXXXL M.

If the sides about an angle of a triangle be unequal, and have
a given ratio to one another, and if the perpendicular from that
angle to the base divide it into segments that have a given ratio
to one another, the triangle is given in species.

Let ABC be a triangle, the sides of which about the angle BAC
are unequal, and have a given ratio to one another, and let the per-
pendicular AD to the base BC divide it into the segments BD, DC,
whidi have a given ratio to one another, the triangle ABC is given
in species. '

Let AB be greater than AC, and make the angle CAE equal to
the angle ABC ; and because the angle AEB is common to the
triangles ABE, CAE, they are (4. 6.) equian- A ^

gular to one another : therefore as AB to BE, ^x^'^^I^V

so Is CA to AE, and by permutation, as AB to ^^y^ i\^
AC, so BE to EA, and so is EA to EC : but B DC E

the ratio of BA to AC is given therefore the ratio of BE to EA, as
also the ratio of EA to EC is given ; where- M

fore (9. dat.) the ratio of BE to EC, as also
(cor. 6. dat) the ratio of EC to CB is given :
and the ratio of BC to CD is given (7. dat)
because the ratio of BD to DC is given;
therefore (9. dat.) the ratio of EC to CD is G K L H B

given, and consequntly (7. dat.) the ratio of

3113 *BUCLID*8 DITA.

D£ to EC: and the ratio of EC to £A was shown to be ^ven, there^
fore (0. dat) the ratio of DE to EA is given ; and ADE is a right
angle« wherefore (46. dat.) the triangle ADE is given in species, and
the angle AED given : and the ratio of CE to EA is given, there^
fore (44. dat.) the triangle AEC is given in species, and conse>-
quently the angle ACE is given, as also the adjacent angle ACB.
la the same manner, because the ratio of BE to EA is given, the
triangle BEA is given in species, and the angle ABE is therefore
given : and the angle ACB is given ; wherefore the triangle ABC is
given (43. dat.) in species.

But the ratio of the greater side BA to the other AC must be less
than the ratio of the greater segment BD to DC: because the
square of BA is to the square of AC, as the squares of BD, DA to
the squares of DC, DA ; and the squares of BD, DA have to the
squares of DC, DA a less ratio than the square of BD has to> the
square of DC,* because the square of BD is greater than the sqcrare
of DC ; therefore the square of BA has to the square of AC a less
ratio than the square of BD has to that of DC : and consequently
tiie ratio of BA to AC is less than the ratio of BD to DC,

This being premised, a triangle which shall have the thmgs
mentioned to be given in the proposition, and to which the triian-
gle ABC is similar, may be found thus: take a straight Ikie GH
given in position and magnitude, and divide it in K, so that the
ratio of GK to EH may be the same with the given ratio of BA
to AC : divide also GH in L, so that the ratio of GL to LH may
be the same with the same ratio of BD to DC, and draw LM at
right angles to GH : and because the ratio of the side of a trian-
gle is less than the ratio of the segments of the base, as has been
shown, the ratio GK to EH is less than the ratio of GL to LH;
wherefore the point L must fall between E and H : also make as
GK to KH ; so GN to NK, and so shall (19. 5.) NK be to NH.
And from the centre N, at the distance NK, describe a eircle,
and let its circumference meet LM in O, and join OG, OH ; then
OGH is the triangle which was to be described : because GN is
to NK, or NO, as NO to NH, the triangle OGN is equiangular
to HON ; therefore as OG to GN, so is HG to ON, and by per-
mutation, as GO to OH, so is GN to NO, or NK, that is, as GK
to KH, that is, in the given ratio of the sides ; and by the construc-
tion, GL, LH have to one another the given ratio of the segments
of the base.

PROP. LXXXII. 6a.

If a parallelogram given in species and magnitude be increas-
ed or diminished by a gnomon given in magnitude, the sides of
the gnomon are given in magnitude.

* If A be grreater than B, and C any third magnitude ; then A and* C together
hMe to B and C together a loss ratio than A has to R

Let A be to B as C to D, and because A is greater than B, C is greater than
D : but as A is to B, so A and C to B and D ; and A and C have to B and C a less ra-
tio than A and C have to B and D, because C is greater than D, there&re A and C
have to B and C a less ratio than A to B.

JBUCLn>>9 DATA* 909

Firat» let the paralleksgraoi AE given in speei«ir mid magnitude be
increased by the given gnomon ECBDFG, each of the straight lihe«
CE, DF is giten.

Because AB is given in species and magnitude, and that the
gnomon E('BDFG is given, therefore ttee whole space AG is
given in magnitude : but AG is also given in species, because it
is similar (2. def.'2. and 24. 6.) to AB; therefore the sides of AG
are given (60. dat) : each of the straight lines
AE, AF 19 therefore given; and each of the
straight lines CA, AD is given (60. dat), there-
fore each of the remainders EC, DP is given
(4. dat.).

Next, let the parallelogram AG given in spe-

cies and magnitude, be diminished by the given [
gnomon ECBDFG, each of the straight lines '
CE, DF is given. ^

Because the paraUelogram AG is given, as also its gnomon
ECBDFG, the remaining space AB is given in magnitude : but it is
also given in species : because it is similar (2. def. 2. and 24. 6.) to
AG; therefore (60. dat.) its sides CA, AD are given, and each of the
straight lines EA, AF is given ; therefore EC, DF are each of them
given.

The gnomon and its sides CE, DF may be found thus in the first
case. Let H be the given space to which the gnomon must be made
equal, and find (25. 6.) a parallelogram similar to AB and equal to
the figures AB and H together, and place its sides AE, AF from the
point A, upon the straight lines AC, AD, and complete the parallelo-
gram AG which is about the same diameter (26. 6.) with AB; be-
cause therefore AG is equal to both AB and H, take away the com-
mon part AB, the remaining gnomon ECBDFG is equal to the re-
maining figure H; therefore a gnomon equal to H, and its sides CE,
DF are found : and in like manner they may be found in the other case,
in which the given figure H must be less than the figure FE from
wiiich it is to be taken.

PROP. LXXXm. 38.

If a parallelogram equal to a given space be applied to a
giveh straight line, deficient by a parallelogram given in species,
the sides of the defect are given.

Let the parallellogram AC equal to a given space be applied
to the given straight line AB, deficient by the parallelogram BDCL
given in species ; each of the straight lines CD, DB are given.

Bisect AB in E; therefore EB is given in magnitude: upon
EB describe (18. 6.) the parallelogram EF similar to DL and si*

Euclid's data.

milariy placed ; therefore EF is given in spe- G HP

cies, and is about the same diameter (26. 6.)

with DL; let BCG be the diameter, and

construct the figure; therefore, because the

figure EF given in species is desdribed upon

the given straight line EB, EF is given (56.

dat.) in magnitude, and the gnomon ELH is

equal (36. and 43. 1.) to the given figure AG : therefore (82. datO

since EF is diminished by the given gnomon ELH, the sides EK, FH

of the gnomon are given ; but EK is equal to DC, and FH to DB ;

wherefore CD, DB are each of them given.

This demcmstration is the analysis of the problem in the 28th
prop, of book 6, the construction and demonstration of which pro-
position is the composition of the analysis ; and because the given
space AC or its equal the gnomon ELH is to be taken from the
figure EF described upon the half of AB similar to BC, therefore AC
must not be greater than EF, as is shown in the 27th prop. B. 6.

PROP.LXXXIV. 5^-

Ir a parallelogram equal to a given space be applied to a
given straight line, exceeding by a parallelogram given m spe-
cies ; the sides of the excess are given.

Let the parallelogram AC equal to a given space be applied
to the given straight line AB, exceeding by the parallelogram
BDCL given in species ; each of the straight lines CD, DB are
given.

Bisect AB in E ; therefore EB is given in magnitude : upon
EB describe (18. 6.) the parallelogram EF similar to LD, and si-
milarly placed ; therefore EF is given in species, and is about the
same diameter (26. 6.) with LD. Let CBG q P H

be the diameter, and construct, the figure :
therefore, because the figure EF given in
species is described upon the given straight
line EB, EF is given in magnitude, (56.
dat.) and the gnomon ELH is equal to the
given figure (36. dat 43. L) AC ; where-
fore since EF is increased by the given gnomon ELH, its sides EEl,
FH are given (82. dat.) ; but EK is equal to CD, and FH to BD ;
therefore CD, DB are each of them given.

This demonstration is the analysis of the problem in the 29th
prop, book 6, the construction and demonstration of which is the
composition of the analysis.

CoR. If a parallelogram given in species be applied to a given
straight line, exceeding by a parallelogram equal to a given space ;
the sides of the parallelogram are given.

Let the parallelogram ADCE given in species be applied to
the given straight line AB, exceeding by the parallelogram BDCXr
equal to a given space ; the sides AD, DC of the parallelogram are
given

Draw the diameter DB of the parallelogram AC, and construct

A —

EUCLID'a DATA. 965

the jSgure. Because the parallelogram AK is equal (48. 1.) to BC
which is given, therefore AK is given ; and E G v C

BK is similar (24. 6.) to AC, therefore BK
is given in species. And since the parallelo-
gram AK given in magnitude is applied to
the given straight line AB, exceeding by the
parallelogram BK given in species, therefore,
by this proposition BD, DK the sides of the A
excess are given, and the straight line AB is given ; therefore the
whole AD, as also DC, to which it has a given ratio, is given.

PROBLEM.

To apply a parallelogram similar to a given one to a given straight
line AB, exceeding by a parallelogram equal to a given space.

To the given straight line AB apply (29. 6.) the parallelogram AK
equal to the given space, exceeding by the parallelogram BK similar
to the one given. Draw DF, the diameter of BK, and through the
point A draw AE parallel to BF, meeting DF produced in E, and
complete the parallelogram AC.

The parallelogram BC is equal (43. 1.) to AK, that is, to the given
space ; and the parallelogram AC is similar (24. 6.) to BK ; therefore
the parallelogram AC is applied to the straight line AB similar to the
one given, and exceeding by the parallelogram BC which is equal to
the given space.

PROP. LXXXV. 84.

Ip two straight lines contain a parallelogram given in magni-
tude, in a given angle ; if the difference of the straight lines be
given, they shall each of them be given.

Let AB, BC contain the parallelogram AC given in magnitude, in
the given angle ABC, and let the excess of BC above AB be given ;
each of the straight lines AB, BC is given.

Let DC be the given excess of BC above BA, A E

therefore the remainder BD is equal to BA. Com- / T

plete the parallelogram AD; and because AB is / /

equal to BD, the ratio of AB to BD is given ; and / /

the angle ABD is given, therefore the parallelo- s^ JT p

gram AD is given in species ; and because the given ^
parallelogram AC is applied to the given straight line DC, exceeding
by the paralldogram AD given in species, the sides of the excess
are given (84. dat.) : therefore BD is given ; and DC is given, where-
fore the whole BC is given : and AB is given, therefore AB, BC are
each of them given.

PROP. LXXXVI. 85.

If two straight lines contain a parallelogram given in magni-
tude, in a given angle ; if both of them together be given, tljusy
shall each of them be given.

3M rtfUCIilC^S D&T'A.

Let the two rtralght Hnes AB, BC contain the parallelogwan AC
given in magnitude, in the given angle ABC, and let AB, BC together
be given ; each of the straight lines AB, BC is given.

Produce CB, and make BD equal to AB, and complete the
parallelogram ABDE. Because DB is equal to BA, and the angle
ABD given, because the adjacent angle ABC is
given, the parallelogram AD is given in spe-
cies; and because AB, BC together are given,
and AB is equal to BD ; therefore DC is given ;
and because the given parallelogram AC is ap-
plied to the given straight line DC, deficient by
the parallelogram AD given in species, the sides AB, BD of the defect
are given (83. dat.) ; and DC is given, wherefore the remainder BC
is given; and each of the straight lines AB, BC is therefore given.

PROP. LXXXVn. 87.

If two straight linos contain a parallelogram given in magni-
tude, in a given angle; if the excess of the square of the greater
above the square of the lesser be given, each of the straight lines
shall be given.

Let the two straight lines AB, BC contain the given parallelo-
gram AC in the given angle ABC ; if the excess of the square of BC
above the square of BA be given, AB and BC are each of them given.

Let the given excess of the ^square of BC above the square of
BA be the rectangle CB, BD : take this from the square of BC ;
the remainder, which is (2. 2.) the rectangle BC CD, is equal to
the square of AB; and because the angle ABC of the parallelo-
gram AC is given, the ratio of the rectangle of the sides AB, BC
to the parallelogram AC is given (62. dat.); and AC is given,
therefore the rectangle AB, BC is given; and the rectangle CB
BD is given; therefore the ratio of the rectangle CB, BD to the
rectangle AB, BC, that is (L 6.), the ratio of the straight line DB
to BA is given ; therefore (54. dat.) the ratio of the square of DB
to the square of BA is given : and the square of
BA is equal to the rectangle BC, CD: where-
fore the ratio of the rectangle BC, CD to the
square of BD is given, as also the ratio of four
times the rectangle BC, CD to the square o^ « pr^
BD; and, by composition (7. dat.), the ratio of ° ^^
four times the rectangle BC, CD, together with the square of BC
to the square of BD is given: but four times the rectangle BC,
CD, together with the square of BD is equal (6. 2.) to the square
of the straight lines BC, CD taken together : therefore the ratio
of the square of BC, CD together to the square of BD is given ;
wherefore (58. dat.) the ratio of the straight line BC, together
with CD to BD, is given: and, by composition, the ratio of BC,
together with CD and DB, that is, the ratio of twice BC to BD,
is given; therefore the ratio of BC to BD is given, as also (1. 6.)
the ratio of the square of BC to the rectangte CB, BD : but the

IIU

£UCLID*8 DATA. 847

rectangle CB, BD is given, being the given excess of the square of
BC, BA ; therefore the square of BC, and the straight line BC, is
given : and the ratio of BC to BD, as also of BD to BA, has been
shown to be given ; therefore (9. dat.) the ratio of BC to BA is given ;
and BO is given, wherefore BA is given.

The preceding demonstration is the analysis of this problem, viz. .

A parallelogram AC which has a given angle ABC being given in
magnitude, and the excess of the square of BC one of its sides above
the square of the other BA being given ; to find the sides : and the
composition is as follows.

Let EFG be the given angle to which the angle ABC is required
to be equal, and from any point E in EF, draw EG perpendicular to
FG; let the rectangle EG, GH be the given j^
space to which the parallelogram AC is to
be made equal; and the rectangle HG,
GL be the given excess of the squares of
BC,BA.

Take, in the straight line GE, GK equal
to FE, and make GM double of GK : join

ML, and in GL produced, take LN equal to ' - ' — ^- „^

LM : bisect GN in O, and between GH, ^ ^ ^ ^ "^^

GO find a mean proportional BC : as OG to GL, so make CB to
BD; and make the angle CBA equal to GFE, and as LG to GK,
so make DB to BA ; and complete the parallelogram AC : AC is
equal to the rectangle EG, GH, and the excess of the squares of CB,
BA is equal to the rectangle HG, GL.

Because as CB to BD, so is OG to GL, the square of CB is to the
rectangle CB, BD as (1. 6.) the rectangle HG, GO to the rectangle
HG, GL : and the square of CB is equal to the rectangle HG, GO»
because GO, BC, GH are proportionals ; therefore the rectangle CB,
BD is equal (14. 5.) to HG, GL. And because as CB to BD, so is
OG to GL ; twice CB is to BD, as twice OG, tliat is GN, to GL:
and, by division, as BC together with CD is to BD, so is NL, that is,
LM, to LG : therefore (22. 6.) the square of BC together with CD is
to the square of BD, as the square of ML to the square of LG : but
the square of BC.and CD together is equal (8. 2.) to four times the
rectangle BC, CD together with the square of BD ; therefore four
times the rectangle BC, CD together with the square of BD is to the
square of BD, as the square of ML to the square of LG: and, by di-
vision, four times the rectangle BC, CD is to the square of BD, as
the square of MG to* the square of GL ; wherefore the rectangle BC,
CD is to the square of BD as (the square of KG the half of MG to
the square of GL, that is, as) the square of AB to the square of BD,
because as LG to GK, so DB was made to BA: therefore (14. 6.)
the rectangle BC, CD is equal to the square of AB. To each of
these add the rectangle CB, BD, and the square of BC becomes
equal to the square of AB together with the rectangle CB, BD ;
therefore this rectangle, that is, the given rectangle KG, GL, is
the excess of the squares of BC, AB. From the point A, diaw
AP perpendicular to BC, and because the angle ABP is equal to

96i eucud'9 data.

the angle KFQt the triangle ABP is equiangular to EFG: and
DB was made to BA, as LQ to GK ; therefore as the rectangle
CB, BD to Ca BA, so is the rectangle HG, GL to HG, GK ;
and as the rectangle CB, BA to AP, BC, so is (the straight

P G L O HN

line BA to AP, and so is FE or GK to EG, and so is) the rectangle
HG, GK to HG, GE; therefore ex sequali, as the rectangle CB,
BD to AP, BC, so is the rectangle HG, GL to EG, GH : and the
rectangle CB, BD is equal to HG, GL: therefore the rectangle
AP, BC, that is, the parallelogram AC, is equal to the given rectan-
gle EG, GH. •

PROP. LXXXVni. N.

If two straight lines contain a parallelogram given in magni-
tude, in a given angle ; if the sum of the squares of its sides be
given, the sides shall each of them be given.

Let the two straight lines AB, BC contain the parallelogram
ABCD given in magnitude in the given angle ABC, and let the sum
of the squares of AB, BC be given ; AB, BC are each of them given.

First, let ABC be a right angle ; and because twice the rectangle
contained by two equal straight lines is equal to both their squares ;

but if two straight lines are unequal, twice the rect- a D

angle contained by them is less than the sum of their ^
squares, as is evident from the 7th prop, book 2,
Elem. ; therefore twice the given space, to which space the rectangle
of which the sides are to be found is equal, must not be greater than
the given sum of the squares of the sides : and if twice that space be
equal to the given sum of the squares, the sides of the rectangle must
necessarily be equal to one another ; therefore in this case describe
a square ABCD equal to the given rectangle, and its sides AB, BC
are those which were to be found : for the rectangle AC is equal to
the given space, and the sum of the squares of its sides AB, BC is
equal to twice the rectangle AC, that is, by the hypothesis, to the
given space to which the sum of the squares was required to be
equal

But if twice the given rectangle be not equal to the given sum
of the squares of the sides, it must be less than it, as has been
shown. Let ABCD be the rectangle; join AC and draw BE
perpendicular to it, and complete the rectangle AEBF, and de-
scribe the circle ABC about the triangle ABC ; AC is its diame-
ter (Cor. 5.4.): and because the triangle ABC is sunilar (8. 6.)
to AEB, as AC to CB, so is AB to BE; therefore the rectangle

BUCUD^fl DATA^ W9

AC, BE is equal to KB, BC ; and the rectangle AB, EC is given,

wherefore AC, BE is given : and because the sum of the squares

of AB, BC is given, the square of AC which is equal (47. 1.) to

that sum is given; and AC itself is therefore given in magnitude:

let AC be likewise given in position, and the point A; therefore

AF is given («32. dat.) in position : and the A D

rectangle AC, BE is given, as has been

shown, and AC is given, wherefore (61.

dat.) BE is given in magnitude, as also

AF which is equal to it ; and AF is also

given in position, and the point A is given ;

wherefore (30. dat.) the point F is given,

and the straight line FB in position (31.

dat) : and the circumference ABC is given

in position, wherefore (28. dat.) the point

& is given : and the points A, C are given ; therefore the straight

lines AB, BC are given, (29. dat.) in position and magnitude.

The sides AB, BC of the rectangle may be found thus : let the
rectangle GH, GK be the given space to which the rectangle
AB, BC is equal; and let GH, GL be the given rectangle to
which the sum of the squares of AB, BC is equal : find (14. 2.)
a square equal to the rectangle GH, GL : and let its side AC be
given in position : upon AC as a diameter describe the semicircle
ABC, and as AC to GH, so make GK to AF, and from the point
A place AF at right angles to AC: therefore the rectangle CA,
AF is equal (16. 6.) to GH, GK; and, by the hypothesis, twice
the rectangle GH, GK is less than GH, GL, that is, than the
square of AC ; wherefore twice the rectangle CA, AF is less than
the square of AC, and the rectangle CA, AF itself less than
half the square of AC, that is, than the rectangle contained by
the diameter AC and its half; wherefore AF is less than the semi-
diameter of the circle, and consequently the straight line drawn
through the point F parallel to AC must meet the circumference
in two points : let B be either of them, and join AB, BC, and com-
plete the rectangle ABCD; ABCD is the rectangle which was to
be found: draw BE perpendicular to AC; therefore BE is equal
(34. 1.) to AF and because the angle ABC in a semicircle is a
right angle, the rectangle AB, BC is equal (8. 6.) to AC, BE,
that is, to the rectangle CA, AF, which is equal to the given
rectangle GH, GK : and the squares of AB, BC are together equal
(47. 1.) to the square of AC, that is, to the given rectangle GH,
GL.

But if the given angle ABC of the parallelogram AC be not
a right angle, in this case, because ABC is a given angle, the
ratio of the rectangle contained by the sides AB, BC to the paral-
lelogram AC is given (62. dat.); and AC is given, therefore the
rectangle AB, BC is given; and the sum of the square of AB,
BC is given; therefore the sides AB, BC are given by the pre-
ceding case.

The sides AB, BC and the parallelogram AC may be fojmd
thus: let EFG be the given angle of the parallelogram, and from

y I

BM EUCLIP'9 DATA.

any point E in F£ draw EG perpendicular to FG; and let the
rectangle EG, FH be the given space to which the parallelogram
is to be made equal, and let EF, FK be the A D

given rectangle to which the sum of the square
, of the sides is to be equal. And, by the pre-
ceding case, find the sides of a rectangle
which is equal to the given rectangle EF, FH, B L C

and the squares of the sides of which are to- ^

gether equal to the given rectangle EF, PK ;
therefore, as was shown in that case, twice
the rectangle EF, FH must not be greater
than the rectangle EF, FK ; let it be so, and
let AB, EC be the sides of the rectangle joined
in the angle ABC equal to the given angle C— _-_!

EFG, and complete the parallelogram ABGD, ^ HG K

which will be that which was to be found : draw AL perpendicular
to BC, and because the angle ABL is equal to EFG, the triangle
ABL is equiangular to EFG; and the parallelogram AC, that is, the
rectangle AL, BC is to the rectangle AB, BC, as (the straight line
AL to AB, that is, as EG to EF, thai is, as) the rectangle EG, FH,
to EF, FH : and, by the construction, the rectangle AB, BC is equal
to EF, FH, therefore the rectangle AL, BC, or, its equal, the paral-
lelogram AC, is equal to the given rectangle EG, FH; and the
squares of AJB, BC are together equal, by construction, to the given
rectangle EF, FK.

PROP. LXXXIX. 86.

If two straight lines contain a given parallelograrp in a given
angle, and if the excess of the square of one of them above a
given space, has a- given ratio to the square of the other ; each
of the straight lines shall be given.

Let the two straight lines AB, BC contain the given parallelogram
AC in the given angle ABC, and let the excess of the square, of BC
above a given space have a given ratio to the square of AB, each of
the straight lines AB, BC is given.

Because the excess of the square of BC above a. given space
has a gii'en ratio to the square of BA, let the rectangle CB, BD
be the given space ; take this from the square of BC ; the remain-
der, to wit, the rectangle (2. 2.) BC, CB has a given ratio to the
square of BA : draw AE perpendicular to BC, and let the square
of BF be equal to the rectangle' BC, CD ; then, because the ang^le
ABC, as also BEA, is given, tlie triangle ABE p

is given (43. dat.) in species, and the ratio of
AE to AB is given : and because the ratio of
the rectangle BC, CD, that is, of the square
of BF to the square of BA is given, the ratio
of the straight line BF to BA is given (58. _ ^

dat.) ; and the ratio of AE to Aft is given, o hsU C
wherefore (9. dat.) the ratio of AE to BF is given ; as also the
ratio of the rectangle AE to BC, that is, (35. I.) of the paral-

BDCLID*S DATA. 361

lelogram AC to the rectangle FB, BC ; and AC is given, wherefore
the rectangle FB, BC is given. The excess of the square of BC
above the square of BF, that is, above the rectangle BC, CD is
given, for it is equal (3. 2.) to the given rectangle CB, BD ; therefore,
because the rectangle contained by the straight lines FB, BC is
given, and also the excess of the square of BC above the square of
BF ; FB, BC are each of them given (87. dat.) ; and the ratio of FB
to BA is given ; therefore, AB, BC are given.

The composition is as follows:

Let GHK be the given angle to which the angle of the paral-
lelogram is to be made equal, and from any point G in HG, draw
GK perpendicular to HK ; let GK, HL be the rectangle to which
the parallelogram is to be made equal, and let
LH, HM be the rectangle equal to the given
space which is to be taken from the square of
one of the sides; and let the ratio of the re-
mainder to the square of the other side be the
same with the ratio of the square of the given HKM L

straight line NH to the square of the given straight line HG.

By help of the 87th dat. find two straight lines BC, BF, which
contain ^ rectangle equal to the given rectangle NH, HL, and such
that the excess of the square of BC above the F

square of BF be equal to the given rectangle /

LH, HM ; and join CB, BF in the angle FBC " M

Ljn, nivi ; ana join ijo, or m me angle voij '/ y

equal to the given angle GHK : and as NH to / /

HG, so make FB to BA, and complete the pa- „— rTf; ^

rallelogram AC, and draw AE perpendicular to " ** *^ ^
BC ; then AC is equal to the rectangle GK, AL ; and if from the
square of BC, the given rectangle LH, HM l>e taken, the remainder
shall have to the square of BA the same ratio which the square of
NH has to the square of HG.

Because, by the construction, the square of BC is equal to the
square of . KF, together with the rectangle LH, HM ; if from the
square of BC there be taken the rectangle LH, HM, there remains
the square of BF, which has (22. 6.) to the square of BA the same
ratio which the square of NH has to the square of HG, because, as
NH to HG, so FB was made to BA ; but as HG to GK, so is BA to
AE, because the triangle GHK is equiangular to ABE ; therefore ex
ssqiudi, ^s NH to GK, so is FB to AE ; wherefore (I. 6.) the rectan-
gle NH, HM is to the rectangle GK, HL, as the rectangle FB, BC to
AE, BC ; but by the construction, the rectangle NH, HL is equal to
FB, BC; therefore (14. 5.) the rectangle GK, HL is equal to the rect-
angle AE, BC, that is, to the parallelogram AC.

The analysis of this problem might have been made as in the 86th
prop, in the Greek, and the composition of it may be made as that
which is in prop. 87th of this edition.

PROP. XC. O.

If two straight lines contain a given parallelogram in a given
angle, and if the square of one of them together with the space

362 Euclid's data.

which has a given ratio to the square of the other be given, each
of the straight lines shall be given.

Let the two straight lines AB, BC contain the given parallelogram
AC in the given angle ABC, and let the square of BC together with
the space which has a given ratio to the square of AB be given ;
AB, BC are each of them given.

Let the square of BD be the space which has the given ratio
to the square of AB; therefore, by the hypothesis, the square of
BC together with the square of BD is given. From the point A,
draw AE perpendicular to BC; and because the angles ABE,
BEA are given, the triangle ABE is given (43. dat.) in species;
therefore the ratio of BA to AE is given ; and because the ratio
of the square of BD to the square of BA is given, the ratio of
the straight line BD to BA is given (58. dat.); and the ratio of
BA to AE is given ; therefore (9. dat.) the ratio of AE to BD is
given, as also the ratio of the rectangle AE, BC, that is, of the
parallelogram AC to the rectangle DB, BC; and AC is given,
therefore the rectangle DB, BC is given; and the square of BC

BE C ^ ^ K

together with the square of BD is given ; therefore (88. dat.) because
the rectangle contained by two straight lines DB, BC is given, and
the sum of their squares is given, the straight line DB, BC are each
of them given ; and the ratio of DB to BA is given : therefore AB,
BC are given.

7%e composition is as follows :

Let FGH be the given angle to which the angle of the parallelo-
gram is to be made equal, and from any point F in GF draw FH
perpendicular to GH ; and let the rectangle FH, GK be that to which
the parallelogram is to be made equal ; and let the rectangle EG, GL
be the space to which the square of one of the sides of the parallelo-
gram together with the space which has a given ratio to the square
of the other side, is to be made equal ; and let this given ratio be the
same which the square of the given straight line MG has to the
square ofGF.

By the 88th dat. find two straight lines DB, BC which contain
a rectangle equal to the given rectangle MG, GK: and such 'that
the sum of their squares is equal to the given rectangle KG, GL :
therefore, by the determination of the problem in that proposi-
tion, twice the rectangle MG, GK, must not be greater than the
rectangle KG, GL. Let it be so, and join the straight lines DB,
BC in the angle DBC equal to the given angle FGH ; and, as MG
to GF, so make DB to BA, and complete the parallelogram AC:
AC is equal to the rectangle Ffa, GK ; and the square of BC to-

Euclid's data.

sea

Zt7

,B E C G H K L

gether with the square of BD, which, by the construction, has to the
square of BA, the given ratio which the square of MG has to the
square of GF, is equal, by the construction, to the given rectangle
KG, GL. Draw AE perpendicular to BO.

Because, as DB to BA, so is MG to GF ; and as BA to AE, so
GF to FH ; ex sequali, as DB to AE, so is MG to FH ; therefore as
the rectangle DB, BC to AE, BC, so is the rectangle MG, GK to FH,
GK ; and the rectangle DB, BC is equal to the rectangle MG, GK ;
therefore the rectangle AE, BC, that is, the parallelogram AC, is
equal to the rectangle FH, GK.

PROP. XCI.

88.

If a straight line drawn within a circle given in magnitude
cuts off a segment which contains a given angle ; the straight
line is given in magnitude.

In the circle ABC given in magnitude, let the straight line AC be
drawn, cutting ofif the segment AEC which contains the given angle
AEG ; the straight line AC is given in magnitude.

Take Q the centre of the circle (1. 3.), join AD and produce it
to E, and join EC : the angle ACE being a
right (31. 3.) angle, is given; and the angle ^
AEC is given ; therefore (43. dat.) the tri-
angle ACE is given in species, and the ra-
tio of EA to AC is therefore given ; and E A
is given in magnitude, because the circle is
given (5. def.) in magnitude ; AC is there-
fore given (2. dat.) in magnitude.

PROP. Xeil. 89.

If a straight line given in magnitude be drawn within a circle
given in magnitude, it shall cut off a segment containing a given
angle.

Let the straight line AC given in magnitude be drawn within the
circle ABC given in magnitude ; it shall cut off a segment containing
a given angle.

Take D the centre of the circle, join AD
and produce it to E, and join EC : and be-
cause each of the straight lines EA and AC
is given, their ratio is given (1. dat.); and
the angle ACE is a right angle, therefore
i:he triangle ACE is given (46. dat.) in spe-
<Aes^ and consequently the angle AEC is
given.

2i64 EUCMD*a DATA.

PROP. XCIIL 90.

Ir from any point in the circumference of a circle given in
position, two straight lines be drawn meeting the circumference,
and containing a given angle ; if the point in which one of them
meets the circumference again be given, the point in which the
other meets it is also given.

From any point A in the circumference of a circle ABC given
in position, let AB, AC be drawn to the cir- A

cumference, making the given angle BAG;
if the point B be given, the point C is also
given.

Take D the centre of the circle, and join
BD, DC; and because each of the points
B, D is given, BD is given (29. dat.) in posi-
tion ; and because the angle BAG is given,
the angle BDC is given (20. 3.), therefore because the straight line
DC is drawn to the given point D in the straight line BD given in
position in the given angle BDC, DC is given (32. dat.) in position :
and the circumference ABC is given in position, therefore (28. dat.)
the point C is given.

PROP.XCIV. 91.

Ir from a given point a straight line be drawn touching a cir-
cle given in position ; the straight line is given in position and
magnitude. ^

Let the straight line AB be drawn from the given point A touch-
ing the circle BC given in position ; AB is given in position and
magnitude.

Take D the centre of the circle, and join DA, DB: because
each of the points D, A is given, the
straight line AD is given (29. dat.) in
position and niEignitude; and DBA is a
right (18. 3.) angle, wherefore DA is a
diameter (Cor. 5. 4.) of the circle DBA,
described about the triangle DBA ; and
that circle is therefore given (6. def.) in
position : and the circle BC is given in
position, therefore the point B is given (28. dat.) ; the point A is also
given : therefore the straight line AB is given (29. dat.) in position
and magnitude.

1>R0P. XCV. 92.

Ir a straight line be drawn from a given point without a circle
given in position ; the rectangle contained by the segments be-
twixt the point and the circumference of the circle is given.

EUCLID* S DATA.

Let the straight line ABC be drawn from the given point A with-
out the circle BCD given in position, cut-
ting it in B, C ; the rectangle BA, AC is
given.

From the point A draw (17. 3.) AD ^
touching the circle ; therefore AD is given \ y 13 yV

(94. dat.) in position and magnitude ; and
because AD is given, the square of AD
is given (56. dat.)i which is equal (36. 8.)

to the rectangle BA, AC: therefore the rectangle BA, AC is
given.

PROP. XCVI. 93.

If a straight line be drawn through a given point within a
circle given in position, the rectangle contained by the seg-
ments betwixt the point and the circumference of the circle is
given.

Let the straight line BAC be drawn through the given point A
within the circle BCE given in position ; the rectangle BA, AC is
given.

Take D the centre of the circle, join AD,
and produce it to' the points E, F; because
the points A, D are given, the straight line
AD is given (29. dat.) in position; and the
circle BEC is given in position ; therefore the
points E, F are given (28. dat.) ; and the point
E is given, therefore EA, AF are each of them
given (29. dat.) ; and the rectangle EA, AF is
therefore given ; and it is equal (35. 3.) to the rectangle BA, AC,
which consequently is given.

. PROP.. XCVII. 94.

1p a straight line be drawn within a circle given in magnitude,
cutting off a segment containing a given angle ; if the angle in
the segment fee bisected by a straight line produced till it meets
the circumference, the straight lines which contain the given
angle shall both of them together have a given ratio to the
straight line which bisects the angle: and the rectangle con-
tained by both of these lines together which contain the given '
angle, and the part of the bisecting line cut below the base of
the segment, shall be given.

Let the straight line BC be drawn within the circle ABC given
in magnitude, cutting off a segment containing the given angle
BAC, and let the angle BAC be bisected by the straight line
AD; BA together with AC has a given ratio to AD; and the
rectangle contained by BA and AC together, and the straiglit

366 EC7CL1D*S DATA.

line ED cut off from AD below BC the base of the segment, is
giyen.

Join BD; and because BC is drawn within the drcle ABC
given in magnitude, cutting off the segment BAG, containing the
given angle BAC; BC is given (91. dat.) in magnitude: by the
same reason BD is given: therefore (1. dat.) the ratio of BC to
BD is given: and because the angle BAC is bisected by AD, as
BA to AC, so is (3. 6.) BE to EG ; and, by permutation, as AB to
BE, so is AC to CE; wherefore (12. 5.) as BA and AC together
to BC, so is AC to CE: and because the angle BAE is equal to
EAC, and the angle ACE to (21. 3.) F
ABD, the triangle ACE is equiangular
to the triangle ABD; therefore as AC
to CE, so is AD to DB: but as AC to
CE, so is BA together with AC to BC :
as therefore BA and AC to BC, so is
AD to DB ; and, by permutation, as BA
and BC to AD, so is BC to BD : and the
ratio of BC to BD is given, therefore the
ratio of BA together with AC to AD is
given.

Also the rectangle contained by BA and AC together, and DE
is given.

Because the triangle BDE is equiangular to the triangle ACE,
as BD to DE, so is AC to CE; and as AC to CE, so is BA and
AC to BC ; therefore as BA and AC to BC, so is BD to DE ;
wherefore the rectangle contained by BA and AC together, and
DE, is equal to the rectangle CB, BD : but CB, BD is given ; there-
fore the rectangle contained by BA and AC together, and DE is
given.

Oihenoise^

Produce C A, and make AP equal to AB, and join BP ; and be-
cause the angle BAC is double (5. and 32. 1.) of each of the angles
BPA, BAD, the angle BPA is equal to BAD ; and the angle BC A is
equal to BDA, therefore the triangle FCB is equiangular to ABD :
as therefore PC to CB, so is AD to DB; and, by permutation, as
PC, that is, BA and AC together, to AD, so is CB to BD : and the
ratio of CB to BD is given, therefore the ratio of BA and AC to AD
is given.

And because the angle BPC is equal to the angle DAC, that is, to
the angle DBC, and the angle ACB equal to the angle ADB, the tri-
angle FCB is equiangular to BDE; as therefore PC to CB, so is BD
to DE ; therefore the rectangle contained by PC, that is, BA and AC
together, and DE, is equal to the rectangle CB, BD, which is given,
and therefore the rectangle contained by BA, AC together, and DE,
is given.

PROP. XCVIII. P.

Ir a straight line be dra^^n within a circle given in magni-
tude, cutting off a segment containing a given angle, if the angle

Euclid's data. 367

adjacent to the angle in the segment be bisected by a straight
line produced till it meet the circumference again and the base
of the segment ; the excess of the straight lines which contain
the given angle shall have a given ratio to the segment of the
bisecting line which is within tne circle ; and the rectangle con-
tained by the same excess and the segment of the bisecting line
betwixt the base produced and the point where it again meets
the circumference, shall be given.

Let the straight line BC be drawn within the circle ABC given in
magnitude, cutting off a segment containing the given angle BAG,
and let the angle CAP adjacent to BAG be bisected by the straight
line DAE, meeting the circumference again in D, and BC the base of
the segment produced in £; the excess of BA, AC has a given ratio
to AD; and the rectangle which is contained by the same excess
and the straight line ED, is given.

Join BD, and through B draw BG parallel to DE meeting AG
produced in G: and because BC cuts off from the circle ABC
given in magnitude the segment j. p

BAG containing a given angle, BG
is therefore given (91. dat.) in mag-]
nitude; by the same reason BD is
given, because the angle BAD is
equal to the given angle EAF:
therefore the ratio of BG to BD is
given : and because the angle CAE
is equal to EAF, of which CAE is
equal to the alternate angle AGB, ^^ ^

and EAF to the interior and opposite angle ABG; therefore the
angle AGB is equal to ABG, and the straight line AB equal to AG ;
so that GG is the excess of BA, AG ; and because the angle BGC is
equal to GAE, that is, to EAF, or the angle BAD ; and that the angle
BCG is equal to the opposite interior angle BDA of the quadrilateral
BCAD in the circle ; therefore the triangle BGC is equiangular to
BDA : therefore as GC to CB, so is AD to DB ; and, by permutation,
as GG which is the excess of BA, AC to AD, so is CB to BD : and
the ratio of CB to BD is given : therefore the ratio of the excess of
BA, AG to AD is given.

And because the angle GBC is equal to the alternate angle
DEB and the angle BCG equal to BDE ; the triangle BCG is
equiangular to BDE : therefore as GC to CB, so is BD to DE ;
and consequently the rectangle GG, DE is equal to the rectangle
CB, BD which is given, because its sides CB, BD are given : there-
fore the rectangle contained by the excess of BA, AG and the straight
line DE is given.

bucud'0 data.

PROP. XCIX.

95.

Ip from a given point in the diameter of a circle given in posi-
tion, or in the diameter producefi, a straight line be drawn to
any point in the circumference, and from that point a straight
line be drawn at right angles to the first, and from the point in
which this meets the circumference again, a straight line be
drawn parallel to the first ; the point in which this paraltel meets
the diameter is given ; and the rectangle contained by the two
parallels is given.

In BC the diameter of the circle ABO given in position, or in
JBC produced, Jet the given point D be taken, and from D let a
straight line DA be drawn to any point A in the circumference,
and let AE be drawn at right angles to DA, and from the point S
where it meets the circumference again let £F be drawn paraUel to
DA meeting BC in F ; the point F is given, as also the rectangle
AD, EF.

Produce EF to the circumference in G, and join AG: because
GEA is a right angle, the straight line AG is (Cor. 5. 4.) the dia-
meter of the circle ABC ; and BC is also a diameter of it ; there-
fore the point H where they meet is the centre of the circle, and
consequently H is given : and the point D is given, wherefore DH
is given in magnitude: and because AD is parallel to FG, and
GH equal to HA; DH is equal (4. 6.) to HF, and AD equal to
GF: and DH is given, therefore HF is given in magnitude; and

it is also given in position, and the point H is given, therefore

(30. dat.) the point F is given. . • *

And because the straight line EFG is drawn from a given point
F without or within the circle ABC given in position, therefore
(95. or 96. dat.) the rectangle EF, FG is given : and EF is ^ual to
AD, wherefore the rectangle AD, EF is given.

PROP. C.

If from a given point in a straight line given in position, a
straight line be drawn to any point in the circumference of a
circle given in position ; and from this point a straight line be ^
drawn, making with the first an angle equal to the difference ot

CUCLID*8 DATA.

300

a right angle and the angle contained by the straight line given
in position, and the straight line which joins the ^iven point and
the centre of the circle; and from the point in wnich the second
line meets the circumference again, a third straight line be drawn,
making with the second, an angle eqpjal to that which the first
makes with the second : the point in which this third line meets
the straight line given in position is given; as also the rectangle
contained by the first straight line and the segment of the third
betwixt the circumference and the straight line given in position,
is given.

Let the straight line CD be drawn from the given point C in
the straight line AB given in position, to the circumference of the
circle DEF given in position, of which G is the centre; join CG,
and from the point D let DF be drawn, making the angle CDF
equal to the difference of a right angle and the angle BCG, and
from the point F let FE be drawn, making the angle DFE equal to
CDF, meeting AB in H : the point H is given ; as also the rectangle
CD, FH.

Let CD, FH meet one another in the
point K, from which draw KL perpendi-
cular to DF ; and let DC meet the circum-
ference again in M, and let FH meet the
same in E, and join MG, GF, GH.

Because the angles MDF, DFE are A
equal to one another, the circumferences
MF, DE are equal (26. 3.) ; and adding or
taking away the common part ME, the
circumference DM is equal to EF ; there-
fore the straight line DM is equal to the
straight line EF, and the angle GMD to
the angle (8. 1.) GFE; and the angles
GMC, GFH, are equal to one another, be-
cause they are either the same with the E
angles GMD, GFE or adjacent to them :
and because the angles EDL, LKD are to-
gether equal (32. L) to a right angle, that
is, by the hypothesis, to the angles KDL»
GCB; the angle GCB, or GCH, is equal
to the angle (LKD, that is, to the angle)
LKF or GHK : therefore the points C, K,
H, G are in the circumference of a circle;
and the angle GCK is therefore equal to A C H B

the angle GHF; and the angle GMC is equal to GFH, and the
straight line GM to GF; therefore (26. 1.) CG is equal to GH, and
CM to HF; Euid because CG is equal to GH, the angle GCH is
equal to GHC; but the angle GCH is given: therefore GHC is

970 soglid's data.

given, and consequently the angle CGH is given ; and OG is given
in position, and the point G; therefore (32. dat.) GH is given in
position ; and CB is also given in position, wherefore the point H
is given.

And because HF is equal to CM, the rectangle DC, FH is equal
to DC, CM : but DC, CM is given (95. or 96. dat), because the point
C is given, therefore the rectangle DC, FR is given.

FINIS.

NOTES ON EUCLID'S DATA.

DEFINITION n.

This is made more explicit than in the Greek text, to prevent a
mistake which the author of the second demonstration of the 24th
proposition in the Greek edition has fallen into, of thinking that a ra-
tio is given to which another ratio is shown to be eqiial, though this
other be not exhibited in given magnitudes. See the Notes on that
proposition, which is the 13th in this edition. Besides, by this defi-
nition, as it is now given, some propositions are demonstrated, which,
in the Greek, are not so well done by help of prop. 2.

DEP. IV.

In the Greek text, de£ 4, is thus ; " Points, lines, spaces, and an*
gles are said to be given in position which have always the same
situation ;" but this is imperfect and useless, because there are in-
numerable cases in which things may be given according to this defi-
nition, and 3ret their position cannot be found ; for instance, let the
triangle ABC be given in position, and let it be proposed to draw
a straight line BD from the angle at B to the opposite side AC,
which shall cut off the angle DBC, which A

shall be the seventh part of the angle
ABC ; suppose this is done, therefore the

straight line BD is invariable in its posi- y^ \ D

tion, that is, has always the same situa- B — :r- . ->.. .. ^ C
tion ; for any other straight line drawn from the point B on eitlier
side of BD cuts ofif an angle greater or lesser than the seventh part
of the angle ABC ; therefore, according to this definition, the strai^t
line BD is given in position, as also (28. dat.) the point D in which it
meets the straight line AC which is given in position. But from
the things here givbn, neither the straight line BD nor the point D
can be found by the help of Euclid's Elements only, by which every
thing in his Data is supposed may be found. This definition is
therefore of no use. We have amended it by adding, "and which,
are either actually exhibited or can be found ;'' for nothing is to be
reckoned given which cannot be found, or is not actually exhibited.

The definition of an angle given by position is taken out of the 4th,
and given more distinctly by itself in the definition marked A,

DEF.XL XII. XIII. XIV. XV.

The 11th and 12th are omitted, because they cannot be given in
English so as to have any tolerable sense : and, therefore, wherever
the terms defined occur, the words which express their meaning are
made use of in their place.

87S iroTn on bucld's data.

The IStfa, 14th, 16th are omitted, as being of no use.

It i« to be observed in general of the Data in this book, that they
are to be understood to be given geometrically, not always arith-
meticaily ; that is, 4hey cannot always be exhibited in nambers ; for
instance, if the side of a square be given, the ratio of it to its diame-
ter is given (4i, dat) geometricaliy, but not in numbers ; and the
diameter is given (2. dat.) ; but though the number of any equai parts
in the side be given, for example, 10, the number of them in the di-
ameter cannot be given : and the like holds in many other cases.

PROPOSITION I.

In this it is shown, that A is to B, as C to D ; from this, that A is
to C, as B to D ; and then by permutation : but it foUows directly,
without these two steps, from 7, 5.

PROP. n.

The limitation added at the end of this proposition between the
inverted commas is quite necessary, because without it the propo-
sition cannot always be demonstrated : for the author having said,*
** because A is given, a magnitude equal to it can be found (1. def.) ;
let this be C; and because the ratio of A to B is given, a ratio
which is the same to it can be found (2. def) ;** adds, *' Let it be
found, and let it be the ratio of C. to A." Now, from the second
definition nothing more follows, than that some ratio, suppose the
ratio of £ to Z, can be found, which is the same with the ratio of A
to B ; and when the author supposes that the ratio of C to A, which
is also the same with the ratio of A to B can be found, he necessa-
rily supposes that to the three magnitudes E, Z, C, a fourth propor-
tional A may be found ; but this cannot always be done by the Ele-
ments of Euclid ; from which it is plain Euclid must have understood
tbe proposition under the limitation which is now added to his text.
An example will make this clear : let A A B

be a given angle, and B another angle
to which A has a given ratio : for in-
stance, the ratio of the given straight
line £ to the given one Z ; then, having
found an angle C equal to A, how can
the angle a be found to which C has the
same ratio that E has to Z ? Certainly /\ ^'

no way, until it be shown how to find
an angle to which a given angle has a
given ratio, which cannot be done by
Euclid's Elements, nor probably by any Geometry known in his
time. Therefore, in all the propositions of this book which depend
upon this second, the above mentioned limitation must be under-
stood, though it be not explicitly mentioned.

PROP. V.

The order of the propositions in the Greek text, between prop.
4, and prop. 25, is now changed into another which is more natu-

* See Dr. 6re|fory*f edition of the Data.

A J9 ^

Norm OM fiociiio's data, 373

ral, by placing those which are more simple before those which are
more complex; and by placing together those which are of the same
kind, some of which were mixed among others of a different kind.
Thus, prop. 12, in the Greek, is now made the 5th, and those which
were the 22d and 28d are made the 11th and 12th, as they are more
simple than the propositions concerning magnitudes, the excess of
one of which above a given magnitude has a given ratio to the other,
after which these two were placed; and the 24th in the Greek text is,
for the same reason, made the 13th.

PROP. VL VII.

These are universally true, though in the Greek text they are
demonstrated by prop. 2, which has a limitation ; they are there-
fore now shown without it.

PROP. xn.

In the 1B3d prop, in the Greek text, which here is the 12th, the
words, " |*Tj <nf « aurx^ 5g," are wrong translated by Claud. Hardy, in
his edition of Euclid's Data, printed at Paris, anno 1 rffes, which was
the first edition of the Greek text: and Dr. Gregory follows him in
translating them by the words, " etsi, non easdem," as if the Greek
had been si xai fi.*] lag avrsg, as in prop. 9. of the Greek text. Euclid's
meaning Is, that the ratios mentioned irl the proposition must not be
the same; for, if they were, the proposition would not be true.
Whatever ratio the whole has to the whole, if the ratios of the parts
of the first to the parts of the other be the same with this ratio, one
part of the first may be double, triple, &c. of the other part of it, or
have any other ratio to it, and consequently cannot have a given ra-
tio to it; wherefore, these words must be rendered by " non autem
easdem," but not the same ratios, as Zambertus has translated them
in his edition.

PROP. xm.

Some very ignorant editor has given a second demonstration of
this proposition in the Greek text, which has been as ignorantly kept
in by Claud. Hardy and Dr. Gregory, and has been retained in the
translations of Zambertus and others ; Carolus Renaldinus gives it
only : the author of it has thought that a ratio was given, if ano-
ther ratio could be shown to be the same to it, though this last ratio
be not found : but this is altogether absurd, because from it would
be deduced, that the ratio of the sides of any two squares is given,
and the ratio of the diameters of any two circles, &c. And it is to
be observed, that the moderns frequently take given ratios, and ra-
tios that are always the same, for one and the same thing ; and Sir
Isaac Newton has fallen into this mistake in the 17th lemma of his
Principia, edit. 1713, and in other places: but this should.be careful*
ly avoided, as it may lead into other errors.

PROP. XIV. XV.

Euclid, in this book, has several propositions concerning mag-
nitudes, the excess of one of which above a given magnitude has

374 NOTEH on soglid^s data.

a given ratio to the other; but he has given none concerning magni-
tudes whereof one together with a given magnKiide has a ^ven ra-
tio to the other; though these last occur as fhsquently in the sdution
of problems as the first; the reason of which is, that the last may
be all demonstrated by help of the first; for, if a magnitude, toge-
ther with a given magnitude, has a given ratio to another magnitude,
the excess of this other above a given magnitude shall have a given
ratio to the first, and on the contrary; as we have demonstrated in
prop. 14. And for a like reason prop. 15 has been added to tlie
Data. One example will make the thing clear : suppose it were to
be demonstrated, that if a magnitude A together with a given mag-
nitude has a given ratio to another magnitude B, that the two magni-
tudes A and B, together with a given magnitude, have a given ratio
to that other magnitude B ; which is the same proposition with re-
spect to the last kind of magnitudes above mentioned, that the first
part of prop. 16, in this edition, is in respect of the first kind: this
is shown thus; from the hypothesis, and by the first part of prop. 14«
the excess of B above a given magnitude has unto A a given ratio ;
and, therefore, by the first part of prop. 17, the excess of B above a
given magnitude has unto B and A together a given ratio ; and by
the second part of prop, 14, A and B together with a given magni-
tude has unto B a given ratio ; which is the thing that was to be
demonstrated. In like manner the other propositions concerning
the last kind of magnitudes may be shown.

PROP. XVI. xvn.

In the third part of prop. 10, in the Greek text, which is the 16th
in this edition, after the ratio of EC to CB has been shown to be
given; from this, by inversion and conversion, the ratio of BO to
BE is demonstrated to be given ; but without these two steps, the
conclusion should have been made only by citing the 6th propo-
sition. And in like manner, in the first part of prop. 11, in the
Greek, which in this edition is the 17th, from the ratio of DB to BO
being given^ the ratio of £>C to DB is shown to be given by inver-
sion and composition, instead of citing prop. 7, and the same fault
occurs in the second part of the same prop. 11,

PROP. XXI. xxn.

These now are added, as being wanting to complete the subject
treated of in the four preceding propositions.

PROP, xxm.

This, which is prop. 20, in the Greek text, was separated from
prop. 14, 15, 16, in that text, after which it should have been Im-
mediately placed, as being of the same kind : it is now put into its
proper place; but prop. 31, in the Greek, is lefl out, as being the
same with prop. 14, in that text, which is here prop. 18.

PROP. XXIV.
This, which is prop. IB, m the Greek, is now put into its proper

NOTS8 OK WCUP^S DATA. 876

place, having been disjoined from the three following It in this edi*
tion, which are of the same kind, .

PROP. XXVIII.

This, which in the Greek text is prop. 25, and, several of the fol-
lowing propositions, are there deduced from def. 4^ which is not
sufficient, as has been mentioned in the note on that definition : they
are therefore now shown more explicitly.

• PROP. XXXIV. XXXVL

Each of these has a determination, which is now added, which
occasions a change in their demonstrations.

PROP. XXXVn. XXXIX. XL. xu.

The 35th and 36th propositions in the Greek text are joined into
one, which makes the 39th in this edition, because the same enuncia-
tion and demonstration serves both : and for the same reason prop.
37, 38, in the Greek, are joined into one, which here is the 40th.

Prop. 87 is added to the Data, as it frequently occurs in the solu-
tion of problems; and prop. 41 is added to complete the rest.

PROP. XLII.

This is prop. 39, in the Greek text, where the whole construction
of prop. 22, of book 1, of the Elements, is put, without need, into the
demonstration, but is now only cited.

PROP. XLV.

This is prop. 42, in the Greek, where the three straight lines made
use of in the con^ruction are said, bxA. not sbown» to be such thai
any I wo of them is greater than the third, which is now done.

PROP. XLVII.

This is prop. 44, in the Greek text ; but the demonstrations of it
is changed into another, wherein the several cases of it are^ shown,
which, though necessary, is not done in the Greek.

PROP. XLVni.

There are two cases in this proposition, arising from the two cases
of the third part of prop. 47, on which the 48th depends ; and in the
composition these two cases are explicitly given.

PROP. LIL

The construction and demonstration of this, which is prop, 48, in
the Greek, are made something shorter than In that text.

PROP. LIIL

Prop. 63, in the Greek text, is omitted, being only a case of prop.
49, in that text, which is prop. 53,. in this edition.

'976 HOTES OH S9CLID*S DATA.

PROP. LVIH.

This is not in the Greek text, bat its demonstration is contained
in that of the first part of prop. 54, in that text ; which proposition is
concerning figures that are given in species : this 68th is true of
similar figures, though they be not given in species, and as it fre-
quently occurs, it was necessary to add it.

PROP.LIX.LXL

This is the 54th in the Greek ; and the 77th in the Greek, being
the very same with it, is left out, and a ^loiter demonstration is
given of prop. 61.

PROP. Lxn.

This, which is most frequently useful, is not in the Greek, and is
necessary to prop. 87, 88, in this edition, as also, though not men-
tioned, to prop. 86, 87, in the former editions. Prop. 66, in the
Greek text, is made a corollary to it.

PROP. LXIV.

This contains both prop, 74, and 73, in the Greek text ; the first

« case of the 74tfa is a repetition of prop. 56, from which it is separated

in that text by many propositions ; and as there is no order in these

propositions, as they stand in the Greek, they are now put into the

order which seemed most convenient and natural.

The demonstration of the first part of prop. 73, in the Greek, is
grossly vitiated. Dr. Gregory says, that the sentences he has
enclosed betwixt two stars are superfluous, and ought to be can-
celled ; but he has not observed, that what follows them is absurd,
being to prove that the ratio [see his figure] of AF to FK is given,
which by the hypothesis at the beginning of the proposition is ex-
pressly given : so that the whole of this part was to be aitered, which
is done in this prop. 64.

PROP. ]LiXvn, ;LXvm.

Prop. 70, in the Greek text, ia divided into these two, for the sake
of distinctness ; and the demonstration of the 67th is rendered shorter
than that of the firsj part of prop. 70, in the Greek, by means of prop.
28, of book 6, of the Elements.

PROP. LXX.

This i» propk 62, in the Greek text ; prop. 78, in that text, is only
a particular case of it, and is therefore omitted.

Dr. Gregory, in the demonstration of prop. 62, cites ^e 49th prop,
dat. to prove that the ratio of the figure AEB to the parallelogram
AH is given ; whereas this was shown a few lines before : . and be-
sides, the 49th prop, is not applicable to these two figures ; because
AH is not given in species, but is, by the step for which the citation
is brought, proved to be given in species.

PROP. Lxxm.

Prop. 13, in the Greek text, is neither well enunciated nor
demonstrated. The 73d, which in tliis edition is put in place of

IfOTES ON eUCLID*S DATA. 377

it, is really the same, a« will appear by considermg [see Dr. Gregory*s
edition] that A, B, F, E in the Greek text are four proportionals; and
that the proposition is to show that A, which has a given ratio to E,
is to r, as B is to a straight line to which A has a given ratio; or,
by inversion, that F is to A, as a straight line to which A has a given
ratio is to B; that is, if the proportionals be placed in this or^er, viz.
r, E, A, B, that the first F is to A to which the second E has a given
ratio, as a straight line to which the third A has a given ratio is to
the fourth B; which is the enunciation of this 73d, and was thus
changed that it might be made like to that of prop. 72, in this edition,
which is the 82d in the Greek text : and the demonstration of prop.
73 is the same with that of prop. 72, only making use of prop. 23,
instead of prop. 22, of book 5, of the Elements.

PROP. Lxxvn.

This is put in place of prop. 70, in the Greek text, which is not a
datum, but a theorem premised as a lemma to prop. 80 in that text :
and prop. 70 is made cor. 1 to prop. 77, in this edition. CI. Hardy,
in his addition of the Data, takes notice, that in prop. 80, of the Greek
text, the parallel KL in the figure of prop. 77, in this edition, must
meet the circumference, but does not demonstrate it, which is done
here at the end of cor. 3, prop. 77, in the construction for finding a
triangle similar to ABC.

PROP. Lxxvm.

The demonstration of this, which is prop. 80, in the Greek, is
rendered a good deal shorter by help of prop. 77.

PROP. LXXIX. LXXX. LXXXI.

These are added to Euclid's Data, as propositions which are often
useful in the solution of problems.

PROP. LXXXU.

This, which is prop. 60, in the Greek text, is placed before the 83d
and 84th, which, in the Greek, are the 58th and 50th, because the
demonstration of these two in this edition are deduced from that of
prop. 82, from which they naturally follow.

PROP. Lxxxvm. XC.

Pr. Gregory, in his preface to Euclid's Works, which he published
at Oxford in 1703, after having told that he had supplied the defects
of the Greek text of the Data in innumerable places from several
manuscripts, and corrected CI. Hardy's translation by Mr. Bernard's;
adds, that the 86th theorem, '* or proposition," seemed to be remark-
ably vitiated, but which could hot be restored by help of the manu-
scripts ; then he gives three different translations of it in Latin, ac-
cording to which, he thinks it may be read ; the two first have no
distinct meaning, and the third which he says is the best, though it
contains a true proposition, which is the 0th in this edition, has no
connexion in the least Vlth the Greek text. And it is strange that
Dr. Gregory did not observe, that, if prop. 86 was changed into this,
the demonstration of the 86th must be cancelled^ and another put in

978 ROTBS Off EUCL10*8 DATA.

its place : but the trath Is, both the enunciation and the demonstra-
tion of prop. 86 are quite entire and right, only prop. 87, which is
more simple, ought to have been placed before it ; and the deficiency
which the doctor justly observes to be in this part of Euclid's Data,
and which, no doubt, is owing to the carelessness and ignorance
of the Greek editors, should have been supplied, not by changing
prop. 86, which is both entire and necessary, but by adding the two
propositions, which are the 88th and 90th in this edition.

PROP. XCVIU. C.

These were communicated to me by two excellent geometers, the
first of them by the Right Honourable the Earl of Stanhope, and the
other by Dr. Matthew Stewart ; to which I have added the demon-
strations.

Though the order of the propositions has been in many places
changed from that in former editions, yet this will be of little disad-
vantage, as the ancient geometers never cite the Data, and the mo-
dems very rarely.

AS that part of the composition of a problem which is its constmc*
tion may not be so readily deduced from the analysis by beginners:
for their sake the following example is given, in which the deviation
of the several parts of the construction from the analysis is particu-
larly shown, that they may be assisted to do the like in other problems.

PROBLEM.

Having given the magnitude of a parallelogram, the angle of which
ABC is given, and also the excess of the square of its side BC above
the square of the side AB; to find its sides, and describe It.
' The analysis of this is the same with the demonstration of the 87th
prop, of the Data, and the construction that is given of the problem
at the end of that proposition is thus derived from the analysis.

Let EFG be equal to the given angle ABC, and because in the
analysis it is said that the ratio of the rectangle AB, BC to the
parallelogram AC is given by the 62d prop, dat., therefore, fit)in
a point in FE, the perpendicular EG is drawn to FG, as the ratio
of FE to EG is the ratio of the rectangle AB, BC to the parallelo-

K
I

^/l

BPDC PGLO HN

gram AC, by what is shown at the end of prop. 62. Next, the

NOTES ON BUCLID's DATA. 879

magnitude of AC is exhibited by making the reetangle BQ, QH
equal to it ; and the given excess of the square of BC above the
square of BA, to which excess the rectangle CB, BL is equal, is
exhibited by the rectangle HG, GL : then, in the , analysis, the
rectangle AB, BC is said to be given, and this is equal to the
rectangle FE, GH, because the rectangle AB, BC is to the paral*
lelogram AC, as (FE to EG, that is, as the rectangle) FE, GH
to EG, GH; and the parallelogram AC is equal to the rectangle
EG, GH, therefore the rectangle AB, BC, is equal to FE, GH:
and consequently the ratio of the rectangle CB, BD, that is, of
the rectangle HG, GL, to AB, BC, that is of the straight line
DB to BA, is the same with the ratio (of the rectangle GU
GH to FE, GH, that is) of tlie straight line GL to FE, which
ratio of DB to BA is the next thing said to be given in the ana-
lysis : from this it is plain that the square of FE is to the square
of GL, as the square of BA. which is equal to the rectangle BC,
CD, is to the square of BD: the ratio of which spaces is the
next thing said to be given: and from this it follows that four
times the square of FE is to the square of GL, as four times the
rectangle BC, CD is to the square of BD; and, by composition,
four times the square of FE together with the square of GL, is
to the square of GL, as four times the rectangle BC, CD, toge-
ther with the square of BD is to the square of BD, that is, (8. 6.)
as the square of the straight lines BC, CD taken together is to
the square of BD, which ratio is the next thing said to be given
in the analysis : and because four times the square of FE and the
square of GL are to be added together ; therefore in the perpen-
dicular EG there is taken KG equal to FE, and MG equal to the
double of it, because thereby the squares of MG, GL, that is,
joining ML, the square of ML is equal to four times the square
of FE and to the square of GL : and because the square of ML
is to the square of GL, as the square of the straight line made up
of BC and CD is to the square of BD, therefore (22. 6.) ML is
to LG, as BC together with CD is to BD ; and, by composition,
ML and LG together, that is, producing GL to N, so that ML
be equal to LN, the straight line NG is to GL, as twice BC is to
BD ; and by taking GO equal to the half of NG, GO is to GL, as
BC to BD, the ratio of which is said to be given in the analysis :
and from this it follows, that the rectangle HG, GO is to HQ,
GL, as the square of BC to the rectangle CB, BD, which is equal
to the rectangle HG, GL ; and therefore the square of BC is equal
to the rectangle HG, GO ; and BC is consequently found by
taking a mean proportional betwixt HG and GO, as is said in
the construction : and because it was shown that GO is to GL, as
BC to BD, and that now the three ifirst are found, the fourth BD
is found by 12. 6. It was likewise shown that LG is to FG, or
GrK, as DB to BA, and the three first are now found, and thereby
the fourth BA. Make the angle ABC equal to EFG, and com-
plete the parallelogram of Vhich the sides are AB, GC, and the
construction is finished ; the rest of the composition contains the
demonstration.

380 NOTia ON CUCLID*S DATA.

As the propositions from the 13th to the 28th may be thought by
beginners to be less useful than the rest, because they cannot ao
readily see how they are to be made use of in the solution of pro-
bienis; on this|iccount the two following problems are added, to show
that they are equally useful with the other propositicms, and from
which it may be easily judged that many other problems depend
upon these propositions.

PROBLEM L

To find three straight lines such, that the ratio of the first to
the second is given ; and if a given straight line be taken from
the second, the ratio of the remainder to the third is given ; also
the rectangle contained by the first and third is given.

. Let AB be the first straight line, CD the second, and EF the
third : and because the ratio of AB to CD is given, and that if a
given straight line be taken from CD, the ratio of the remainder
to EF is given : therefore (24. dat*) the excess of the first AB
above a given straight line has a given ratio to the third EF ; let
BH be that given straight line ; therefore AH, the
excess of AB above it, has a given ratio to EF ; A H B

and consequently (L 6.) the rectangle BA, AH, 1

has a given ratio to the rectangle AB, EF, which C G D

last rectangle is given by the hypothesis; there- 1

fore (2. dat) the rectangle BA, AH is given, and E F

BH the excess of its sides is given ; wherefore the

sides AB, AH are given (85. dat.) : and because K N M L O

the ratios of AB to CD, and of AH to EF are \—\—\

given, CD and EF are (2. dat.) given.

The Composition.

Let the given ratio of KL to KM be that which AB is required to
liave to CD ; and let DG be the given straight line which is to be
taken from CD, and let the given ratio of KM to KN be that which
the remainder must have to EF; also let the' given rectangle NK,
KO be that to which the rectangle AB, EF is required to be equal :
find the given stright line BH which is to be taken from AB, which
is done, as plainly appears from prop. 24. dat. by making as KM to
KL, so GD to HB. To the given straight line BH apply (29. 6)
a rectangle equal to LK, KO exceeding by a square, and let BA, AH
be its sides : then is AB the first of the straight lines requured to be
found, and by making as LK to KM, so AB to DC, DC will be the
second : and lastly, make as KM to KN, so CG to EF, and EF is
the third.

For as AB to CD, so is HB to GD, each of these ratios being
the same with the ratio of LK to KM ; therefore (19. 5.) AH is to
CG, as (AB to CD, that is, as) LK to KM ; and as CG to EF, so
is KL to KN: wherefore, ex sequali, as AH to EF, so is LK to
KN: and as the rectangle BA, AH to the rectangle BA, EF, so is
(1. 6.) the rectangle LK, KO to the rectangle KN, KO: and by the
construction, the rectangle BA, AH is equal to LK, KO: there-

NOTES ON Euclid's data. * 381

fore (14. 5.) the rectangle AB, EF is equal to the given rectangle
NK, KO : and AB has to CD the given ratio of KL to KM ; and from
CD the given straight line GD being taken, the remainder C6 has to^
EF the given ratio of EM to EN. Q,. E. D.

PROB. n.

To find three straight lines such, that the ratio of the first to
the second is given ; and if a given straight line be taken from
the second, the ratio of the remainder to the third is given ; also
the sum of the squares of the first and third is given.

Let AB be the first straight line, BC the second, and BD the third :
and because the ratio of AB to BC is given, and that if a given
straight line be taken from BC, the ratio of the remainder to BD is
given ; therefore (24. dat.) the excess of the first AB above a given
straight line, has a given ratio to the third BD : let AE be that given
straight line, therefore the remainder EB has a given ratio to BD ;
let BD be placed at right angles to EB, and join DE ; thqp the trian-
gle EBD is (44. dat.) given in species ; wherefore the angle BED is
given : let AE, which is given in magnitude, be given also in posi-
tion, as also the point E, and the straight line ED will be given (32.
dat.) in position : join AD, and because the sum of the squares of
AB, BD, that is (47. 1.), the square of AD is given, therefore the
straight line AD is given in magnitude ; and it is also given (34. dat.)
in position, because from the given point A it is drawn to the straight
line ED given in position : therefore the point D, in which the two
straight lines AD, ED given in position cut one another, is given
(28. dat.) : and the straight line DB which is at right angles to AB is
given (33. dat.) in position, and AB is given in position, therefore
(28. dat.) the point B is given : and the points A, D are given, where-
fore (29. dat.) the straight lines AB, BD are given ; and the ratio of
AB to BC is given, and therefore (2. dat) BC is given.

The Composition.

Let the given ratio of FG to GH be that which AB is required to
have to BC, and let HK be the given straight line which is to be taken
from BC, and let the ratio which the remainder is required to have

L
D

A EBNM CFG H K

to BD, be the given ratio of HG to LG, and place GL at right angles
to FH, and join LP, LH : next, as HG is to GP, so make HK to
AE; produce AE to N, so that AN be the straight line to the square
of which the sum of the squares of AB, BD is required to be equ&l ;

389 NOTIB OH BU0UD*8 DATA.

and make the angle NED equal to the angle GFL; and from the cen-
tre A at the distance AN describe a circle, and let its circumference
meet ED in D, and draw DB perpendicular to AN, and DM, making
the angle BDM equal to the angle GLH. Lastly, produce BM to C,
so that MC be equal to HE ; then is AB the first, BC the second, and
BD the third of the straight lines that were to be found.

For the triangles EBD, FGL, as also DBM, LGH being equiangu-
lar, as EB to BD, so is FG to GL ; and as DB to BM, so is LG to
GH ; therefore, ex ssquali^ as EB to BM, so is (FG to GH, and so is)
AE to HK or MC ; wherefore (12. 5.) AB is to BC, as AE to HE,
that is, as FG to GH, that is, in the given ratio ; and from the straight
line BC taking MC, which is equal to the given straight line HE, the
remainder BM has to BD the given ratio of HG to GL ; and the sum
of the squares of AB, BD is equal (47. L) to the square of AD or AN,
which is the given space. Qi. E. D.

I believe it would be in vain to try to deduce the preceding con-
struction from an algebraical solution of the problem.

rrois.

THB

ELEMENTS

PLANE AND SPHERICAL

TRIGONOMETRY.

PLANE TRIGONOMETRY.

LEMMA L Fia 1.

Lbt ABC be a rectilineal angle ; If about tbe point B a« a centre,
and with any distance BA, a circle be described^ meeting BA, BG,
the straight lines including the angle ABC in A, C ; the angle ABC
will be to four right angles, at the arch AC to the whole circum*
ference. [ \

Produce AB till it meet the circle again in F, and through B draw
DE perpendicular to AB, meeting the circle in D, E.

By 83. 6. EIem« the angle ABC is to a right angle ABO, as the
arch AC to the arch AD; and quadrupling the consequents, the
angle ABC will be to four right angles, as the arch AC to four tknes
the arch AD, or to the whole circumference.

LEMMA II. FIG. 2.

Let ABC be a plane rectilineal angle as before : about B as a cen-
tre, with any two distances BD, BA, let two circles be described
meeting BA, BC in D, E, A, C ; the arch AC will be to the whole
circumference of which it is an arch, as the arch DE is to the whole
<^rcumference of which it is ah arch.

By Lemma 1, the arch AC is to the w^de circumferenoe of which
it is an arch, as the angle ABC is to four rigi)t angles ; and by the
same Lemma 1, the arch DE is to the whde circumference of whieh
it is an arch, as the angle ABC is to four right angles ; therefore the
arch AC is to the whole circumference of which it is an arch, as the
arch DE to the whole circumference of which it is an ardi.

DEFINITIONS. FIG. 3.

Lbt ABC be a plane rectilineal angle; if about B as a centre,
with BA any distance, a circle ACF be described, meeting BA,
BC in A, C; the arch AC is called the measure of the angle
ABC.

11.

The circumference of a circle is supposed to be divided into
800 equal parts called degrees ; and each degree into 60 equal
pcuts called minutes, and each minute into 60 equal parts called
seconds, &a And as many degrees, minutes, seconds, &c, as
are contained in any arch, of so many degrees, minutes, seconds,
&>o, is the angle, of which that arch is the measure, said to be.

CoR. Whatever be tbe radius of the circle of which the measure
of a given angle is an arch, that arch will contain the same
number of degrees, minutes, seconds, ioc. as is manifest fbom
Lemma 2.

380 PLANE TRIGONOMETRY.

III.

Let AB be produced till it meet the circle again in F ; the angle CBF,
which, together with ABC, is equal to two right angles, is called
the Supplement of the angle ABC.

IV.

A straight line CD drawn through C, one of the extremities of the
arch AC perpendicular upon the diameter passing through the
other extremity A, is called the Sine of the arch AC, or of the
angle ABC, of which it is the measure.

CoR. The Sine of a quadrant, or of a right angle, is equal to the
radius.

The segment DA of the diameter passing through A, one extremity
of the arch AC, between the sine CD and that extremity, is called
the Versed Sine of the arch AC, or angle ABC.

VI.

A straight line AE, touching the circle at A, one extremity of the
arch AC, and meeting the diameter BC passing through the other
extremity C in E, is called the Tangent of the arch AC ; or of the
angle ABC.

vn.

The straight line BE, between the centre and the extremity of
the tangent AE, is called the Secant of the arch AC, or angle
ABC.

CoR. to def. 4, 6, 7. The sine, tangent, and secant of any angle
ABC, are likewise the sine, tangent, and secant of its supplement
CBF.

It is manifest from def 4, that CD is the side of the angle CBF.
Let CB be produced till it meet the circle again in O ; and it is
manifest that AB is the tangent, and BE the secant, of the angle
ABG or EBF, from def. 6, 7.

Cor. to def. 4,. 5, 6, 7. The sine, versed sine, tangent, and secai;it, of
any arch which is the measure of any given angle ABC, is to the
sine, versed sine, tangent and secant, of any other arch which is
the measure of the same angle, as the radius of the first is to the
radius of the second.

Let AC, MN be measures of the angle ABC, according to def 1,
CD the sine, DA the versed line, AE the tangent, and BE the
secant of the arch AC, according to def 4, 5, 6, 7, and NO the
sine, OM the versed line, MP the tangent, and BP the secant
of the arch MN, according to the same definitions. Since CD,
NO, AE, MP are parallel, CD is to NO as the radius CB to
the radius NB, and AE to MP as AB to BM, and BC or BA to
BD as BN or BM to BO; and, by conversion, DA to MO as
AB to. MB. Hence the corollary is manifest; therefore, if the
'radius be supposed to be divided into any given number of
equal parts, the sine, versed sine, tangent, and secant, of any

PLANE TRIOONOMBTRY. 387

given angle, will each contain a given number of these parts ; and,
by trigonometrical tables, the length of the sine, versed sine, tangent,
and secant, of any angle, may be found in parts of which the radius
contains a given number ; and, vice versa, a number expressing the
length of the sine, versed sine, tangent, and secant, being given, the
angle of which it is the sine, versed sine, tangent, and secant, may be
found.

VIII. Pig. 3.

The difference of an angle from a right angle is called the comple-
ment of that angle. Thus, if BH be drawn perpendicular to AB,
the angle CBH will be the complement of the angle ABC, or of
CBP.

IX.

Let HE be the tangent, CL or DB, which is equal to it, the sine and

BE the secant of CBH, the complement of ABC, according to def.

4, 6, 7, HE is called the cotangent, BD the eo-«me, and BE the

cO'Secant, of the angle ABC.
CoR. 1. The radius is a mean proportional between the tangent and

co-tangent.
For, since HE, BA are parallel, the angles HEB, ABC will be equal,

and the angles EHB, BAE are right : therefore the triangles BAE,

EHB are similar, and therefore AE is to AB, as BH or BA to HE.
CoR. 2. The radius is a mean proportional between the co-sine and

secant of any angle ABC.
Since CD, AE are parallel, BD is to BC or BA, as BA to BK

PROP. I. PIG. 5.

In a right angled plane triangle, if the hypothenuse be made
radius, the sides become the sines of the angles opposite to them :
and if either side be made radius, the remaining side is the tan*
gent of the angle opposite to it, and the hypothenuse the secant
of the same angle.

Let ABC be a right angled triangle ; if the hypothenuse BC be
made radius, either of the sides AC will be the sine of the angle ABC
opposite to it ; and if either side BA be made radius, the other side
AC will be the tangent of the angle ABC opposite to it, and the hy-
pothenuse BC the secant of the same angle.

About B as a centre, with BC, BA for distances, let two circles CD,
EA be described meeting BA, BC in D, E : since CAB is a right an-
gle, BC being radius, AC 4s the sine of the angle ABC by def 4, and
BA being radius, AC is the tangent, and BC the secant of the angle
ABC, by def. 6, 7.

CoR. L Of the hypothenuse a side and an angle of a right angled
triangle, any two being given, the third is also given.

CoR. 2. Of the two sides and an angle of a right angled triangle,
any two being given, the third is also given.

386 FLAI\B TRIGONOMCTRT.

PROP. U. FIG. 6, 7.

The sides of a plane triangle are to one*another as the sines
of the angles opposite to them.

In right angled triangles, this prop, is manifest from prop. I ; for
if the hypothenuse be made radius, the sides are the sines of the an-
gles opposite to them, and the radias is the sine of a right angle (cor.
to def. 4.) which is opposite to the hypothenuse.

In any oblique angled triangle ABC, any two sides AB, AC will be
to one another as the sines of the angles ACB, ABC which are oppo-
site to them.

From C, B draw CE, BD perpendicular upon the opposite sides
AB, AC produced, if need be. Since CEB, CDB, are right angles,
BC being radius, CE is the sine of the angle CBA, and BD the sine
of the angle ACB : but the two triangles CAE, DAB have each a
right angle at D and E; and likewise the common angle CAB;
therefore they are similar, and consequently, CA is to AB as CE
to DB; that is, the sides are as the sines of the angles opposite to
them.

Cor. Hence of two sides, and two angles opposite to them, in a
plane triangle, any three being given, the fourth is also given.

PROP. IIL FIG. 8.

Iif a plane trianglp, the sum of any two sides is to their differ-
ence, as the tangent of half the sum of the angles at the base, to
the tangent of half their difference.

Let ABC be a plane triangle ; the sum of any two sides, AB, AC
will be to their difference as the tangent of half the sum of the an-
gles at the base ABC, ACB to the tangent of half their difference.

About A as a centre, with AB the greater side for a distance, let
a circle be described, meeting AC produced in E, F, and BC in D ;
join DA, EB, FB : and draw FG parallel to BC, meeting EB in G.

The angle EAB (32. 1.) is equal to the sum of the angles at the
base, and the angle EFB at the circumference is equal to the half of
EAB at the centre (20. 3.) ; therefore EFB is half the sum of the
angles at the base ; but the angle ACB (32. I .) is equal to the an-
gles CAD and ADC, or ABC together : therefore FAD is the differ-
ence of the angles at the base, and FBD at the circumference, or BFG,
on account of the parallels FG, BD, is the half of that difference; but
since the angle EBF in a semicircle is a right angle (I. of this), FB
being radius, BE, BG, are the tangents of the angles EFB, BFG; but
it is manifest that EC is the sum of the sides BA, AC, and CF their
difference ; and since BC, FG are parallel (2. 6.), EC is to CF, as
EB to BG ; that is, the sum of the sides is to their difference, as the
tangent of half the sum of the angles at the base to the tangent of
half their difference.

PLANE TRIGOMOMfiTRY. 889

PROP. IV. FIG. 18.

In any plane triangle BAC, whose two sides are BA, AC, and
base BC, the less of the two sides, which let be BA is to the
greaterAC, as the radius is to the tangent of an angle; and the
radius is to the tangent of the excess of this angle above half a
right angle, as the tangent of half the sum of the angles B and
C at the base is to the tangent of half their difference.

At the point A, draw the straight line EAD perpendicular to BA :
make AE, AP each equal to AB, and AD to AC ; join BE, BF, BD,
and from D draw DG perpendicular upon BF. And because BA is
at right angles to EF, and EA, AB, AF are equal^ each of the angles
EBA, ABF is half a right angle, and the whole EBF is a right angle ;
also (4. l/El.) EB is equal to BF. And since EBF, FGD are right
angles, EB is parallel to GD, and the triangles EBF, FGD are simi-
lar ; therefore EB is to BF as DG to GF, and EB being equal to BF,
FG must be equal to GD. And because BAD is a right angle, BA
the less side is to AD or AC the greater, as the radius is to the
tangent of the angle ABD ; and because BGD is a right angle, BG is
to GD or GF as the radius is to the tangent of GBD, which is the
excess of the angle ABD above ABF half a right angle. But
because EB is parallel to GD, BG is to GF as ED is to DF, that is,
since ED is the sum of the sides BA, AC, and FD their difference
(3. of this), as the tangent of half the sum of the angles B, C, at the
base, to the tangent of half their difference. Ttterefore, in any plane
triangle, &c. Q. E. D.

PROP. V. FIG. 9. 10.

In any triangle, twice the rectangle contained by any two
sides is to the difference of the sum of the squares of these two
sides, and the square of the base, as the radius is to the co-sine
of the angle included by the two sides.

Let ABC be a plane triangle; twice the rectangle ABC contained
by any two sides BA, BC, is to the difference of the sum of the
squares of BA, BC, and the square of th*^ base AC, as the radius to
the co-sine of the angle ABC.

From A, draw AD perpendicular upon the opposite side BC ; then
(by 12. and 13. 2. EL) the difference of the sum of the squares of
AB, BC, and the square of the base AC, is equal to twice the rect-
angle CBD ; but twice the rectangle CBA is to twice the rectanf?te
CBD, that is to the difference of the sum of the squares of AB, BC,
and the square of AC (1. 6.), as AB to BD ; that is, by prop. I , as
radius to the sine of BAD, which is the complement of the angle
ABC, that is, as radius to the co-sine of ABC.

390 PLANE TRIGONOMETRY.

PROP. VI. FIG. 11.

In any triangle ABC, whose two sides are AB, AC, and base
EC, the rectangle contained by half the perimeter, and the excess
of it above the base BC, is to the rectangle *containe4 by the
straight lines by which the half of the perimeter exceeds the
other two sides AB, AC, as the square of the radius is to the
square of the tangent of half the angle BAC opposite to the
base.

Let the angles BAC, ABC be bisected by the straight lines AG,
EG; and producing the side AB, let the exterior angle CBH be
bisected by the straight line BK, meeting AG in K ; and from the
points G, K, let there be drawn perpendicular upon the sides the
straight lines GD, GE, GF, KH, KL, KM. Since therefore (4. 4.)
G in the centre of the circle inscribed in the triangle ABC, GD,
GF, GE will be equal, and AD will be equal to AE, BD to BF, and
CE to CF; in like manner KH, KL, KM will be equal, and BH
will be equal to BM, and AH to AL, because the angles HBM,
HAL are bisected by the straight lines BK, KA: and because in
the triangles KCL, KCM, the sides LK, KM are equal, KC is com-
mon, and KLC, KMC are right angles, CL will be equal to CM :
since therefore BM is equal to BH, and CM to CL; BC will be
equal to BH and CL together ; and, adding AB and AC together,
AB, AC, and BC will together be equal to AH and AL together :
but AH, AL are equal: wherefore each of them is equal to half
the perimeter of the triangle ABC: but since AD, AE are equal,
and BD, BF, and also CE, CF, AB, together with FC, will be equal
to half the perimeter of the triangle to which AH or AL was
shown to be equal; taking away therefore the common AB, the
remainder FC will be equal to the remainder BH ; in the same man-
ner it is demonstrated, that BF is equal to CL; and since the
points B, D, G, F, are in a circle, the angle DGF will be equal to
the exterior and opposite angle FBH (22. 3.); wherefore their
halves BGD, HBK will be equal to one another : the right angled
triangles BGD, HBK will therefore be equiangular, and'GD will
be to BD, a.s BH to HK, and the rectangle contained by GD, HK
will be equal to the rectangle DBH or BFC ; but since AH is to
HK, as AD to DG, the rectangle HAD (22. 6.) will be to the rect-
angle contained by HK, DG, or the rectangle BFC, (as the square
of AD is to the square of DG, that is) as the square of the radius
to the square of the tangent of the angle DAG, that is, the half of
BAC : but HA is half the perimeter of the triangle ABC, and AD
is the excess of the same above HD, that is, above the base BC :
but BF or CL is the excess of HA or AL above the side AC ; and
FC, or HB, is the excess of the same HA above the side AB ; there-
fore the rectangle contained by half the perimeter, and the excess
of the same above the base, viz. the rectangle HAD, is to the rect-
angle contained by the straight lines by which the half of the peri-
meter exceeds the other two sides, that is, the rectangle BFC, as

PLANE TRIGONOMETRY. 891

the square of the radius is to the square of the tangent of half the
angle BAG opposite to the base. Q. E. D.

PROP. VII. FIG. 12. 13.

Iir a plane triangle, the base is to the suoi of the sides, as the
i^iflference of the sides is to the sum or difference of the segments
of the base made by the perpendicular upon it from the vertex,
according as the square of the greater side is greater or less
than the sum of the squares of the lesser side and the base.

Let ABC be a plane triangle ; if from A the vertex be drawn a
straight line AD perpendicular upon the base BC, the base BC will
be to the sum of the sides BA, AC, as the difference of the same
sides is to the sum or difference of the segments CD, BD, according
as the square of AC the greater side is greater or less than the sum
of the squares of the lesser side AB, and the base BC.

About A as a centre, with AC the greater side for a distance, let
a -circle be described meeting AB produced in E, F, and CB in G:
it is manifest, that FB is the sum, and BE the difference of the
sides ; and since AD is perpendicular to GC, GD, CD will be equal ;
consequently GB will be equal to the sum or difference of the seg-
ments CD, BD, according as the perpendicular AD/ meets the base,
or the base produced ; that is, (by con v. 12. 13. 2.) according as the
square of AC is greater or less than the sum of the squares of AB,
BC: but (by 35. 3.) the rectangle CBG is equal to the rectangle
EBF; that is, (16. 6.) BC is to BF, as BE is to BG, that is, the base
is to the sum of the sides, as the difference of the sides is the sum
or difference of the segments of the base made by the perpendicular
from the vertex, according as the square of the greater side is
greater or less than the sum of the squares of the lesser side and
the base. Q. E. D.

PROP. VHL PROB. FIG. 14.

The sum and difference of two magnitudes being given, to
find them.

Half the given sum added to half the given difference, will be the
greater, and half the difference subtracted from half the sum, will be
the less.

For, let AB be the given sum, AC the greater, and BC the less.
Let AD be half the given sum ; and to AD, DB, which are equal, let
DC be added ; then AC will be equal to BD and DC together ; that
is, to BC, and twice DC ; consequently twice DC is the difference,
and DC half that difference ; but AC the greater is equal to AD, DC ;
that is, to half the sum added to half the difference, and BC the less
is equal to the excess of BD, half the sum above DC half the differ-
ence. Q, E. D.

992 PLANB TRIOONOMBTRY.

SCHOLIUIB.

Of the six parts of a plane triangle (the three sides and three
angles) any three being given, to find the other three is the business
of plane trigonometry ; and the several cases of that problem may be
resolved by means of the preceding proposition, as in the two follow-
ing, with the tables annexed. In these, the solution is expressed by
a fourth proportional to three given lines; but if the given parts^
be expressed by numbers from trigonometrical tables, it may be
obtained arithmetically by the common Rule of Three.

Note. In the tables, the Ibllowingf abbreviations are useft: R is put for the
Radios; T for Tangent; and S for Sine. Degrees, minutes, seconds, &e. are
written in this manner ; 30^ 25' 13'', &.C., which signifies 30 degrees, 25 minutes,
13 seconds, &c.

PLANS TRIGONOMflTRY.

SOLUTION OF TH£ CaSeS OF RIGHT ANGLED TRIANGLES.

GENERAL PROPOSITION.

In a right angled triangle, of the three sides and three anglers,
any two being given besides the right angle, the other three may
be found, except when the two acute angles are given; in which
case the ratios of the sides are only given, being the same with
the ratios of the sines of the angles opposite to them.

It is manifest from 47, 1, that if of the two sides and hypothe-
nuse any two be given, the third may also be found. It is also
manifest from 32, 1, that if one of the acute angles of a right angled
trismgle be given, the other is also given, for it is the complement of
the former to a right angle.

If two angles of any triangle be given, the third is also given,
being the supplement of the two given angles to two right an-
gles.

The other cases' may be resolved by help of the preceding propo-
sitions, as in the following table :

GIVEN.

SOUGHT.

Two sides, AB, AC.

The an-
gles B, C.

AB : AC : : R : T, B, of which C is
the complement

AB, BC, a side and
the hypothenuse.

AB, B, a side and an
angle.

The an-
gles B, C.

BC : BA : : R : S, C, of which B is
the complement

The other
side AC.

R : T, B : : BA : AC.

AB and B, a side and
an angle.

The hypo-
thenuse BC.

S, C : R : : BA : BC.

BC, and B, the hypo-
thenuse and an angle.

The side
AC.

R : S, B : ; BC ; CA.

These five cases are resolved by prop. 1.

304

PLANE TRIOOMOMSTRY.

SOLUTION OF THE CASES OF OBLIQUE ANGLED TRIANGLES.

GENERAL PROPOSITION.

In an oblique angled triangle, of the three sides and three
angles, any three being given, the other three may be found,
except when theihree angles are given; in which case the ratios
of the sides are only given, being the same with the ratios of
the sines of the angles opposite to them.

GIVEN.

SOUGHT.

A, B, and therefore
C, and the side AR

AB, AC, and B, two

sides and an angle oppo-
site to one of them.

BC,AC.

The an-
gles A and

S, C : S, A : : AB : BC, and also,
S, : S, B : : AB : Aa (2.) (Fig.
16. 17.)

AC : AB : : S, B : S, C, (2.) This
case admits of two solutions; for C
may be greater or less than a quad
rant. (Cor. to def. 4.)

AB, AC, and A, two

sides and the included
angle.

AB+AC : AB— AC : : T, C+B :
The an
gles B andT, C — B : (8.) the sum and difierence

of the angles C, B bein? given each
of them is given. (7.) Otkmmse. Fig.
18.

BA:AC::R:T,ABC,andal8oR:
T, ABC-iS*^ : T, B+C : T, B— C :

2 8

(4.) therefore B and C are given as
before. (7.)

PLAMB TRIGONOMBTRT.

395

OIYBN.

BOUGHT.

AB, BC, CA, the
three sides.

A, B, C,

the three
angles.

2 AC X CB : ACo + CB^—
ABa : : R : Co S, C. If ABq

+ CBq be greater than AB^. Fig.
16.

2 AC X CB : AB^— AC^—
CB^-h : : R :*;o S, C. If AB^i
be greater than ACqxCBq, Fig.
17. (4.)

Othertoise,
Let AB+BC + AC =2 P. P +

• ■

P— AB : P— AC + P—BC
R^ : Tq, i C, and hence C is
known. (5.)

Otkenoise,
Let AD be perpendicular to BC.
1. If ABq be less than AC^+CBg.
Fig. 16. BC : BA + AC : : BA—
AC : BD— DC, and BC the sum of
BD, DC is given ; therefore each of
them is given. (7.)

2. If AB^ be greater than AC^
-fCBo. Fig. 17. BC:BA+AC::
BA— AC : BD+DC ; and BC the
difierence of BD, DC is given,
therefore each of them is given. (7.)

And CA : CD : : R : Co S,C. (1.)
and C being found, A and B are
found by case 2, or 3.

SPHERICAL TRIGONOMETRY.

DEFLMTIONS.

The pole of a circle of the sphere is a point in the superficies of
the sphere, from which all straight , lines drawn to the circumfer-
ence of the circle are equal.

A great circle of the sphere is any whose plane passes through the
centre of the sphere, and whose centre therefore is the same with
that of the sphere.

ni.

A spherical triangle is a figure upon the superficies of a sphere
comprehended by three arches of three great circles, each of
which is less than a semicircle.

IV.

A spherical angle is that which on the superficies of a sphere is con-
tained by two arches of great circles, and is the same with the
inclination of the planes of these great circles.

PROP. I.
Great circles bisect one another.

As they have a common centre, their common section will be a
diameter of each which will bisect them.

PROP. IL FIG. 1.

The arch of a great circle betwixt the pole and the circumfer-
ence of another is a quadrant

Let ABC be a great circle, and D its pole; if a great circle
DC pass through D, and meet ABC in C, the arch DC will be a
quadrant.

Let the great circle CD meet ABC again in A, and let AC be
the common section of the great circles which will pass through
E the centre of the sphere: join DE, DA, DC: by def. 1, DA
DC are equal, and AE, EC are also equal, and DE is common;
therefore (8. 1.) the angles DEA, DEC are equal; wherefore the
arches DA, DC are equal, and consequently each of them is a
quadrant, d. E. D.

PROP. m. FIG. 2.

If a great circle be described meeting two great circles AB,
AC passing through its pole A in B, C, the ansles at the centre
of the sphere upon the circumference BC, is the same with the

SPHERICAL TRIOONOMBTRT, 397

spherical angle BAG, and the arch BC is called the measure of
the spherical angle BAG.

Let the planes of the great circles AB, AC intersect one another
in the straight line AD passing through D their common centre;
join DB, DC.

Since A is the pole of BC, AB, AC will be quadrants, and the
angles ADB, ADC right angles; therefore (6. def. 11.), the angle
CDB is the inclination of, the planes of the circles AB, AC : that is>
(def. 4.) the spherical angle BAC. d. E. D.
* CoR. If through the point A, two quadrants AB, AC be drawn,
the point A will be the pole of the great circle BC, passing through
their extremities B, C,

Join AC, and draw AE a straight line to any other point E in
BC; join DE: since AC, AB are quadrants, the angles ADB, ADC
are right angles, and AD will be perpendicular to the plane of BC :
therefore the angle ADE is a right angle, and AD, DC are equal to
AD, DE, each to each ; therefore AE, AC are equal, and A Is the pole
of BC by def. 1.' Q. E. D.

PROP. IV. FIG. 3.

In isosceles spherical triangles, the angles at the base are
equal.

Let ABC be an isosceles triangle, and AC, CB the equal side?; the
angles BAC, ABC, at the base AC, are equal.

Let D be the centre of the sphere, and join DA, DB, DC; in
DA take any point E, from which draw, in the plane ADC, the
straight line EF at right angles to ED, meeting CD in F, and
draw, in the plane ADB, EG at right angles to the same ED;
therefore the rectilineal figure FEG is (6. def. 11.) the inclination
of the planes ADC, ADB, and therefore is the same with the
spherical angle BAC: from F draw FH perpendicular to DB,
and from H draw, in the plane ADB, the straight line HG at
right angles to HD, meeting EG in G, and join GF. Because
DE is at right angles to EF and EG, it is perpendicular to the
plane FEG (4. 11.), and therefore the plane FEG, is perpendicu-
lar to the plane ADB, in which DE is (18. 11.): in the same man-
ner the plane FHG is perpendicular to the plane ADB, and there-
fore GF the common section of the planes FEG, FHG, is per-
pendicular to the plane ADB (19. 11.); and because the angle
FHG is the inclination of the planes BDC, BDA, it is the same
with the spherical angle ABC ; and the sides AC, CB of the sphe-
rical triangle being equal, the angles EDF, HDF, which stand
upon thenr at the centre of the sphere, are equal ; and in the tri-
angles EDF, HDF, the side DF is common, and the angles DEF,
DHF are right angles ; therefore EF, FH are equal, and in the
triangles FEG, FHG the side GF is common, and the sides EG, GH
will be equal by the 47. 1. and therefore the angle FEG is equal to
FHG (8. 1.) ; that is, the spherical ahgle BAC is equal to the sphe-
rical angle ABC.

898 BPHBRICAL TRIGONOMfiTRr.

PROP. V. FIG. 3.

Ir, in a spherical triangle ABC, two of the angles BAC,
ABC, be equal, the sides BC, AC, opposite to them, are equal.

Read the construction and demonstration of the preceding pro-r
position, unto the words " and the sides AC, CB," &c. and the rest
of the demonstration will be as follows, viz.

And the spherical angles BAC, ABC, being equal, the recti-
lineal angles FEG, FHG, which are the same with them, are
equal ; and in the triangles FGE, FGH the angles at G are right
angles, and the side FG opposite to two of the equal angles is
common; therefore (26. 1.) EF is equal to FH; and in the right
angled triangles DEF, DHF the side DF is common; wherefore
(47. 1.) ED is equal to DH, and the angles EDF, HDF, are therefore
equal (4. 1 .), and consequently the sides AC, BC of the spherical
triangle are equal.

PROP. VI. FIG. 4.

Any two sides of a spherical triangle are greater than the
third.

Let ABC be a spherical triangle, any two sides AB, BC will be
greater than the other side AC.

Let P be the centre of the sphere ; join DA^ DB, DC.

The solid angle at D, is contained by three plane angles ADB,
ADC, BDC; and, by 20. 11. any two of them ADB, BDC are great-
er than the third ADC ; that is, any two sides AB, BC of the spher-
ical triangle ABC, are greater than the third AC.

PROP. Vn. FIG. 4.

The three sides of a spherical triangle are less than a circle.

Let ABC be a spherical triangle as before, the three sides AB, BC,
AC, are less than a circle.

Let D be the centre of the sphere : the solid angle at D is con-
tained by three plane angles BDA, BDC, ADC, which together
are less than four right angles (21. II.); therefore the sides AB,

BC, AC together, will be less than four quadrants: that is, less
than a circle.

PROP. VIII. FIG. 5.

• lif a spherical triangle the greater angle is opposite to the
greater side; and conversely.

Let ABC be a spherical triangle, the greater angle A is opposed to
the greater side BC.

Let the angle BAD be made equal to the angle B, and then

BD, DA will be equal, (5. of this,) and therefore AD, DC are
equal to BC, but AD, DC, are greater than AC, (6. of this),
therefore >BC is greater than AC, that is, the greater angle A is

SPHBIUCAI. TRIGONOMETRT. 3M

opposite to the greater side BC. The converse is demonstrated as
prop. 19, 1. El. Q. E. D.

PROP. IX. FIG. 6.

In any spherical triangle ABC, if the sum of the sides AB,
BC, be greater, equal, or less than a sennicircle, the internal
angle at the base AC will be greater, equal, or less than the ex-
ternal or opposite BCD ; and therefore the sum of the angles A
and ACB will be greater, equal, or less, than two right angles.

Let AC, AB produced meet in D.

1. If AB, BC be equal to a semicircle, that is, to AD, BC, BD will
be equal, that is (4. of this), the angle D, or the angle A will be equal
to the angle BCD.

2. If AB, BC together be greater than a semicirde, that is,
greater than ABD, BC will be greater than BD; and therefore
(8. of this) the angle D, that is, the angle A, is greater than the
angle BCD.

3. In the same manner it is shown, that if AB, BC together be
less than a semicircle, the angle A is less than the angle BCD*
And since the angles BCD, BCA are equal to two right angles*
if the angle A be greater than BCD, A and ACB together will
be greater than two right angles. If A be equ^l to BCD, A and
ACB together will be equal to two right angles; and if A be
less than BCD, A and ACB will be less than two right angles.
Q. E. D.

PROP. X. FIG. 7.

Ir the angular points, A, B, C, of the spherical triangle ABC
be the poles of three great circles, these great circles by their
intersections will form another triangle PDE, which, is called
supplemental to the former ; that is, the sides FD, DE, EF are
the supplements of the measures of the opposite angles C, B, A
of the triangle ABC, and the measures of the angles F, D, E of
the triangle FDE, will be the supplements of the sides, AC, BC,
BA in the triangle ABC.

Let AB produced meet DE, EF in G, M, and AC meet FD, FE in
K, L, and BC meet FD, DE in N, H.

Since A is the pole of FE, and the circle AC passes through A,
EF will pass through the pole of AC (13. 15. 1. Th.), and since
AC passes through C, the pole of FD, FD will pass through the pole
of AC ; therefore the pole of AC is in the point F, in which the
arches DF, EF intersect each other. In the same manner, D is the
pole of BC, and E the pole of AB.

And since F, E are the poles of AL, AM, FL and EM are
quadrants, and FL, EM together, that is, FE and ML together,

400 SPHERICAL TRIQONOMBTRT.

are equal to a semicircle. But since A is the pole of ML, ML is the
measure of the angle BAG, consequently FE is the supplement of
the measure of Ae angle BAG. In the same manner, £D, DF are
the supplements of the measures of the angles ABG, BGA.

Since likewise GN, BH are quadrants, GN, BH together, that is
NH, BG together are equal to a semicircle ; *and since D is the pole
of NH, NH is the measure of the angle FDE, therefore the measure
of the angle FDE is the supplement of the side BG. In the sacne
manner, it is shown that the measures of the angles DEF, EFD are
the supplements of the sides AB, AG in the triangle ABC. Ct S* D«

PROP. XL FIG. 7,

The thre^ angles of a spherical triangle are greater than two
right angles, and less than six right angles.

The measures of the angles A, B, G, in the triangle ABG, together
with the three sides of the supplemental tnan^e DEF, are (10. of
this) equal to three semicircles ; but the three sides of the triangle
FDE, are (7. of this) less than two semicircles ; therefore the mea-
sures of the angles A, B, G are greater than a semicircle; and hence
the angles A, B, G are greater than two right angles.

All the external and internal angles of any triangle are equal to
six right angles ; therefo^je all the internal angles are less than six
right angles.

PROP. Xn. FIG. 8.

It from any point C, which is not the pole of the great circle
ABD, there be drawn arches of great circles CA, CD, CE»
CF, &c., the greatest of these is CA, which passes through H
the pole of A^D, and CB the remainder of ACB is the least,
and of any others CD, CE, CF, &c. CD, which is nearer to CA,
is greater than CE, which is more remote.

Let the common section of the planes of the great circles AGB,
ADB be AB ; and from G, draw CG perpendicular to AB, whioh will
also be perpendicular to the plane ADB (4. def. 11.) ; join GD, GE,
GF, GD, GE, GF, GA, GB.

Of all the straight lines drawn from G to the circumference ADB*
GA is the greatest, and GB the least (7. 3.) ; and GD, which is nearer to
GA, is greater than GE, which is more remote. The triangles GGA»
CGD are right angled at G,and they have the common side CG: there-
fore the squares GG, GA together, that is, the square of GA, is greater
than the squares of GG,GD together, that is, the square of CD; and
GA is greater than GD, and therefore the arch GA is greater than CD.
In the same manner, since GD is greater than GE, and GE than GF,
&c. it is shown that GD is greater than CE, and G£ than GF, &c. ; an^

8PHSRICAL TRIOOirOMETRT. 401

consequently the arch CD greater than the arch CE; and the
arch CE greater than the arch CF, &;c. And since GA is the
greatest, and GB the least of all the straight lines drawn from G
to the circumference ADB, it is manifest that CA is the great-
est, and CB the least of all the straight lines drawn from C to the
circumference: and therefore the arch CA is the greatest, and
CB the least of all the circles drawn through C, meeting ADB.
CI. E.D.

PROP. XIII. FIG. 9.

Iir a right angled spherical triangle, the sides are of the same
aflfection with the opposite angles ; that is, if the sides be greater
or less than quadrants, the opposite angles will be greater or less
than right angles.

Let ABC be a spherical triangle right angled at A ; any side AB,
will be of the same affection with the opposite angle ACB.

Case 1. Let AB be less than a quadrant, let AE be a quadrant, and
let 'EC be a great circle passing through E, C. Since A is a right
angle, and AE a quadrant, E is the pole of the great circle AC, and
£CA a right angle : but ECA is greater than BCA, therefore BCA is
less than a right angle. Q,. E. D.

Fig. 10. — Case 2. Let AB be greater than a quadrant, make AE a
quadrant, and let a great circle pass through C, E ; ECA is a right
angle as before, and BCA is greater than ECA, that is, greater than
a right angle. Q. E. D.

*
PROP. XIV.

If the two sides of a right angled spherical triangle be of the
same affection, the hypothenuse will be less than a quadrant:
and if they be of different afTection, the hypothenuse will be
greater than a quadrant.

Let ABC be a right angled spherical triangle ; if the two sides
AB, AC be of the same or of different affection, the h3rpothenuse
BC will be less or greater than a quadrant.

Fig. 9. — Case 1. Let AB, AC be each less than a quadrant.
Let AE, AG be quadrants; G will be the pole of AB, and E the pole
of AC, and EG a quadrant; but by prop. 12, CE is greater than CB,
since CB is farther off from CGD than CK In the same manner, it
is shown that CB, in the triangle CBD, where the two sides CD, BD
are each greater than a quadrant, is less than CE, that is, less than
a quadrant. Q. E. D.

Fig. 10. — Case 2. Let AC be less, and AB greater than a quad-
rant; then the hypothenuse BC will be greater than a quadrant; for
let^AE be a quadrant, then E is the pole of AC, and EC will be a
quadrant. But CB is greater than CS by prop. 12, since AG passes
through the pole of ABD. Q. E. D.

409 0f M9EICAL TRIGOKOMBTRT*

PROP. XV.

If the hypotbenuse of a right angled triangle be greater or
less than a quadrant, the sides will be of different or the same
affection.

This is the converse of the preceding, and demonstrated in the
same manner.

PROP. XVI.

In any spherical triangle ABC, if the perpendicular AD from
A on the base BC, fall within the triangle, the angles B and C at
the base will be of the same affection ; and if the perpendicular
fall without the triangle, the angles B and C will be of different
affection.

Fig. 11.— 1. Let AD fall within the triangle; then (13. of this),
since^ADB, ADC are right angled spherical triangles, the angles B,
C, must each be of the same affection as AD.

Fig. 12.— 2. Let AD fall without the triangle ; then (13. of thia),
the angle B is of the same affection as AD ; and by the same the
angle ACD is of the same affection as AD ; therefore the angles
ACB and AD are of different affection, and the angles B and ACB
of different affection.

CoB. Hence if the angles B and C be of the same affection, the
perpendicular will fall within the base ; for, if it did not (16. of ihis\
B and C would be of different affection. And if the angles B and
C be of opposite affection, the perpendicular will fall without the
triangle; for, if it did not (16. of this), the angles B and C would
be of the same affection, contrary to the supposition.

PROP. XVII. FIG. 13.

In right angled spherical triangles, the sine of either of the
sides about the right angle, is to the radius of the sphere* as the
tangent of the remaining side is to the tangent of the angle op-
posite to that side.

Let ABC be a triangle, having the right angle at A ; and let AB
be either of the sides ; the sine of the side AB will be to the radius,
as the tangent of the other side AC to the tangent of the ai^le
ABC, opposite to AC. Let D be the centre of the sphere; join
AD, BD, CD, and let AE be drawn perpendicular to BD, which
therefore will be the sine of the arch AB, and from the point E, let
there be drawn in the plane BDC the straight line EF at right an*
gles to BD, meetiiig DC in F, and let AF be joined. Since there*
Ibre the straight line DE is at right angles to both £A and £F, it
will also be at right angles to the plane AEF (4« 11.) ; wherefiore
the plane ABD, which passes through DE, is perpendicular to the
plane AEF (18^ 1).) and the plane AEF perpendicular to ASD ;
the plane ACD or AFD is also perpendicular^ to the- same ABD :

SPHERICAL TIUOONOMITRTr 403

■

therefore the oommon section, viz. the straight line AF, is at right
angles to the plane ABD (19. ll.)» and FAE, FAD are right angles
(3. def. 11.); therefore AF is the tangent of the arch AC ; and in the
rectilineal triangle AEF, having a right angle £lt A, AE will be to
tlie radius as AF to the tangent of the angle AEF (1. PI. Tr.); but
AB is the sine of the arch AB, and AF the tangent of the arch AC,
and the angle AEF is the inclination of the planes CBD, ABD (6.
def. 11.), or the spherical angle ABC: therefore the sine of the arch
AB is to the radius^ as the tangent of the arch AC, to the tangent of
the opposite angle ABC.

Cor. 1. If therefore of the two sides, and an angle opposite to
one of them, any two be given, the third will also be given.

CoR. 2. And since by this proposition the sine of the side AB is
to the radius, as the tangent of the other side AC to the tangent of
the angle ABC opposite to that side; and as the radius is to the co-
tangent of the angle ABC, so is the tangent of the same angle ABC
to the radius (Cor. 2. def. Pi. Tr.) ; by equality, the sine of the side
AB is to the co-tangent of the angle ABC adjacent to it, as the tan-
gent of the other side AC to the radius.

PROP. XVm. FIG. 13.

Iw right angled spherical triangles, the sine of the hypothenuse
is to the radius, as the sine of either side is to the sine of the
luigle opposite to that side.

Let the triangle ABC be right angled at A, and let AC be either
of the sides ; the sine of the hypothenuse BC will be to the radius,
ds the sine of the arch AC is to the sine of the angle ABC.

Let D be the centre of the sphere, and let CG be drawn perpendi-
cular to DB, which will therefore be the sine of the hypothenuse BC ;
and from the point G let there be drawn in the plane ABD the
straight line GH perpendicular to DB, and let CH be joined ; CH
will be at right angles to the plane ABD, as was shown in the pre-
ceding proposition of the straight line FA ; wherefore CHD, CHG
are right angles, and CH is the sine of the arch AC ; and in the
triangle CHG, having the right angle CHG, CG is to the radius, as
CH to the sine of the angle CGH (I. PI. Tr.); but since CG, HG are
at right angles to DBG, which is the common section of the planes
CBD, ABD, the angle CGH will be equal to the inclination of these
planes (6. def. 11.); that is to the spherical angle ABC. The sine
therefore of the hypothenuse CB is to the radius, as the sine of the
side AC is to the sine of the opposite angle ABC. CI. E. D.

CoR. Of these three, viz. the hypothenuse, a side, and the angle
opposite to that side, any two being given, the third is also given
by prop. 2.

PROP. XIX FIG. 14.

I«r right angled spherical triangles, the co-sine of the hypothe-
nuse is to the radius, as the co-tangent of either of the angles i«
to the tangent of the remaining angle.

#04 fTHBRlCAL TRIOONOVBTRT.

' Let ABC be a spherical trian^, having a right apgie at A; the
co-sine of the hypothenuse BC will be to the radius, as the co-tangent
of the angle ABC to the tangent of the angle ACB.

Describe the circle DE, of which B is the pole, and let it meet AC
in F, and the circle BC in E; and since the circle BD passes through
the pole B of the circle DF, DF will also pass through the pole BD
(13. 18. 1. Theod. sph.). And since AC is perpendicular to BD, AC will
also pass through the pole of BD ; wherefore the pole of the circle
BD will be found in the point where the circles AC, DE meet, that is,
!n the point F : the arches FA, EF are therefore quadrants, and like-
wise the arches BD, BE t in the triangle CEF, right angled at the
point E, CE is the complement of the hypothenuse. BC of the triangle
ABC, BF is the complement of the arch ED, which is the m.easure
of the angle ABC, and FC the hypothenuse of the triangle CEF, is
the complement of AC ; and the arch AD, which is the measure of
the angle CFE, is the complement of AR

But (17. of this) in the triangle CEF, the sine of the side CE is to
the radius, as the tangent of the other side is to the tangent of the
angle ECF opposite to it, that is, in the triangle ABC, the co-sine of
the hypothenuse BC is to the radius, as the co-tangent of the angle
ABC is to the tangent of the angle ACB. QL E. D.

Cor. 1. Of these three, viz. the hypothenuse and the two angles^
any two being given, the third will also be given.

Cor. 2. And since by this proposition the co-sine of the hypothe-
nuse BC is to the radius as the co-tangent of the angle ABC to the
tangent of the angle ACB ; but as the radius is to the co-tangent of
the angle ACB, so is the tangent of the same to the radius (Cor. 2.
def. PI. Tr.) ; and ex aequo, the co-sine of the hypothenuse BC is to
the co-tangent of the angle ACB, as the co-tangent of the angle ABC
to the radius.

PROP. XX. FIG. 14.

In right angled spherical triangles, the co-sine of an angle is
to the radius, as the tangent of the side adjacent to that angle is
to the tangent of the hypothenuse*

The same construction remaining; in the triangle CEF (17. of
this), the sine of the side EF is to the radius, as the tangent of the
other side CE is to the tangent of the angle CFE opposite to it; that
is, in the triangle ABC, the co-sine of the angle ABC is to the radius,
as (the co-tangent of the hypothenuse BC to the co-tangent of the side
AB, adjacent to ABC, or as) the tangent of the side AB to the tangent
of the hypothenuse, since the tangents of two arches are reciprocally
proportional to their co-tangents. (Cor. 1 def. PI. Tr.)

Cor. And since by this proposition the co-sine of the angle ABC
is to the radius, as the tangent of the side AB is to the tangent of
the hypothenuse BC ; and as the radius is to the co-tangent of EC,
so is the tangent of BC to the radius; by equality, the co-sine of the
angle ABC will be to the co-tangent of the hypothenuse BC, as the
tangent of the side AB, adjacent to the angle ABC, to the radius.

•PHSJUOAL TUOQNOtlfiTBr.

PROP. XXI. FIG. 14.

409

In right angled spherical triangles, the co-sine of either of the
sides is to the radius, as the co-sine of the hypothenuse is to the
co-sine of the other side.

The same construction remaining ; in the triangle CEF, the sine
of the hypothenuse CF is to the radius, as the sine of the side CE
to the sine of the opposite angle CFE (18. of this); that is, in the
triangle ABC, the co-sine of the side CA is to the radius, as the
co-sine of the hypothenuse BC to the co-sine of the other side BA.
Q. £. D.

PROP. XXn. FIG. 14.

In right angled spherical triangles, the co-sine of either of the
sides is to the radius, as the co-sine of the angle opposite to that
tide is to the sine of the other angle.

The same construction remaining ; in the triangle CEF, the sine
of the hypothenuse CF is to the radius, as the sine of the side
EF is to the sine of the angle ECF opposite to it ; that is, in the
triangle ABC, the co-sine of the side CA is to the radius, as the
cosine of the angle ABC opposite to it, is to the sine of tlie other
angle, d. B. D.

406 anamtGAh TBi«oiroiuT»r;

OF THE CIRCULAR PARTS.

Fig. 16.— Ik any right angled spherical triangle ABC, the com-
plement of the hypothenuse, the complements of the angles, and
the two sides, are called The circular parts of the triangle, as if it
were following each other in a circular order, from whatever part
we begin : thus, if we begin at the complement of the hypothenuse,
and proceed towards the side BA, the parts following in order
will be the complement of the hypothenuse, the complement of the
angle B, the side BA, the side AO, (for the right angle at A is not
reckoned among the parts,) and, lastly, the complement of the
angle C. And thus at whatever part we begin, if any three of
these five be taken, they either will be all contiguous or adjacent,
or one of them will not be contiguous to either of the other two :
in the first case, the part which is fc>etween the other two is called
the Middle part, and the other two are called Adjacent extremes.
In the second case, the part which is not contiguous to either of
the other two is called the Middle part, and the other two OppO'
site extremes. For example, if the three parts be the complement
of the hypothenuse BC, the complement of the angle B, and the
side BA ; since these three are contiguous to each other, the com-
plement of the angle B will be the middle part, and the comple-
ment of the hypothenuse BC and the side BA will be adjacent ex-
tremes: but if the complement of .the hypothenuse BC, and the
sides BA, AC be taken; since the complement of the hypothenuse
is not adjacent to either of the sides, viz. on account of the com-
plements of the two angles B and C intervening between it and
the sides, the complement of the hypothenuse BC will be the mid-
dle part, and the sides BA, AC opposite extremes. The most acute
and ingenious Baron Napier, the inventor of Logarithms, contrived
the two following rules concerning these parts, by means of which
all the cases of right angled spherical triangles are resolved with the
greatest ease.

RULE I.

The rectangle contained by the radius and the sine of the middle
part, is equal to the rectangle contained by the tangents of the
adjacent parts.

RULE n.

The rectangle contained by the radius and the sine of the middle
part, is equal to the rectangle contained by the co-sines of the
opposite parts.

These rules are demonstrated in the following manner :
Fig. J 5. First. Let either of the sides, as BA, be the middle part,
and therefore the complement of the angle B, and the side AC will
be adjacent extremes. And by Cor. 2. prop. 17, of this, S, BA is to
the Co-T, B, as T, AC is to the radius, and therefore B x S,
BA = Co.T, BxT. AC.

The same side BA, being the middle part, the complement of
the hypothenuse, and the complement of the angle C, are opposite

9PSBBJC4L THWOIfOlfBTRy.

407

extromesf ; and by propL 18. S« B C ie to the radius, as S, BA to S, C;
therefore R x S, BA = S, BC x S, C.

Secondly, Let the complement of one of the angles, as B, be the
middle part, and the comi^ement of the h3rpothenuse, and the side
BA will be adjacent extremes: and by Cor. prop. 20. Co-S, B is
to Co-T, BC, as T, BA, is to the radius, and therefore R x Co-S,
B=Co.T,BCxT,BA.

Again, Let the complement of the angle B be the middle part, and
the complement of the angle C, and the side AC will be opposite
extremes ; and by prop. 22, Co-S, AC is to the radius, as Co-S, B is
to S, C : and therefore R x Co-S, B=Co-S, AC x S, C.

Thirdly, Let the complement of the hypothenuse be the middle
part, and the complements of the angles B, C, will be adjacent
extremes : but by Cor. 3. prop. 19, Co-S, BC is to Co-T, B, as Co-T,
C to the radius : therefore R x Co-S, BC = Co-T, C x Co-T, A.

Again, Let the complement of the hypothenuse be the middle part,
and the sides AB, AC, will be opposite extreoaes : but by prop: ^1.
Co-S, AC is to the radius, as Co-S, BC, to Co-S, BA; therefore
R x Co-S, BC = Co-S, BA x Co-S, AC. O. K D.

406

flPHBEICAL THIGONOHBTRT.

tOLUTION or TBI SIXTEEN CASES OF RIGHT ANGLED SPHERICAL TRI-
ANGLES.

GENERAL PROPOSITION.

In a right angled spherical triangle, of the three sides and
three angles, any two being given, besides the right angle, the
other three may be found.

In the following Table the solutions are derived from the preceding
propositions. It is obvious that the same solutions may be de>
rived from Baron Napier's two rules above demonstrated, which,
as they are easily remembered, are commonly used in practice.

Case.

Given.

So't.

AC,C

R : Co-S, AC : : S, C : Co-S, B : and B is of
Uie same species witli CA, by 22, and 13.

AC,B

Co^, AC : R : : Co-S, B : S, C : by 22.

B.C

S, C : Co-S, B : : R : Co-S, AC : by 22. and
AC is of the same species with B. 18.

BA,AC

R : Co-S, BA : : Co-S, AC : Co-S, BC. 21. and
if both BA, AC be grearer or less than a quad-
rant, BC will be less than a quadrant But if
they be of different affections, BC will be greater
than a quadrant 14.

BA,BC

Co-S, BA : R : : Co-S, BC : Co-S, AC, 21.
and if BC be greater or less than a quadrant,
BA, AC will be of different or the same afiec-
tion: by 15.

BA,AC

S, BA : R : : T^ CA : T, B, 17, and B is of the
same affection with AC, 18.

BA, B

R : S, BA : : T,B : T, AC. 17. And AC is

of the same affection with B. 18.

AC,B

T, B : R : : T, CA : S, BA. 17.

SPHERICAL TRIGONOMBTRY.

409

Case

Given.

So't

jBC| C

•

R : Co-S, C : : T, BC : t, CA. 20. If BC be

lessior greater than a quadrant, C and B will be
of the same or different affection. 15. 13.

•

AC,C

•

Co^, C : R : : T, AC : T, BC. 20. And BC

is less or greater than a quadrant, a9Cordinff as
C and AC or C and B are of the same or differ-
ent affection. 14. 1.

BC,CA

T, BC : R : : T, CA : Co^,C. 20. If BC be

less or greater than a quadrant, CA and AB, and
therefore CA and C, are of the same or different
affection. 15.

BCy B

R : S, BC ; : S,*B : S, AC. 18. And AC is
of the same affection with B.

AC,B

•S, B : S, AC : : R : S, BC. 18.

BC,AC

S, BC : R : : S, AC : S, B. 18. And B isof
the same affection with AC.

B,C

T, C : R : : Co-T, B : Co^, BC. 19. And ac-
cording as the angles B and C are of different or
the same afl^ction, BC will be greater or less
than a quadrant. l4

BC,C

R : Co-S, BC : : T, C : Co-T, B. 19. If BC
be less or greater than a quadrant, C and B will
be of the same or different affection. 15.

The second, eighth, and thirteenth ca^es, which are commonly
called ambiguous, admit of two solutions: for in these it is not
determined whether the side or measure of the angle sought be
greater or less than a quadrant. ^

PROP. XXHL FIG. 16.

Iir spherical triangles, whether right angled or oblique angled,
the sines of the sides are proportional to t^e sines of the angles
opposite to them.

First, Let ABC be a right angled triangle, having a right angle
at A; therefore by prop. 18, the sine of the hypothenuse BC is to
the radius (or the sine oi the right angle at A) as the sine, of the
side AC to the sine of the angle B. And in like manner, the sine

410 SPHERICAL TRIOONOMETRY.

of BC is to the sine of the angie A as the sine of AB to the sine of
the angle C ; wherefore (11.5.) the sine of the side AC is to the sine
of the angle B, as the sine of AB to the sine of the angle C.

Secondly, Let BCD be an oblique angled triangle, the sine of either
c^the sides BC, will be to the sine of either of the other two CD, as
the sine of the angle D opposite to BC is to the sine of the angle B
opposite to the side CD. Through the point C, let there be drawn
an arch of a great circle CA perpendicular upon BD ; and in the
right angled triangle ABC (18. of this) the sine of BC is to the radius,
as the sine of AC to the sine of the angle B ; and in the triangle
ADC (by 18. of this) : and, by inversion, the radius is to the sine of
DC as the sine of the angle D to the sine of AC : therefore, ex aequo
perturbate, the sine of BC is to the sine of DC, aa the sine of the
angle D to the sine of the angle B» €1. E. D.

PROP. XXIV. FIG. 17. 18.

In obliqae angled spherical triangles, having drawn a perpen-
dicular arch from any of the angles upon the opposite side, the
co-sines of the angles at the base are proportional to the sines of
the vertical angles.

Let BCD be a triangle, and the arch CA perpendicular to* the base
BD ; the co-sine of the angle B will be to the co-sine of the angle D, as
the sme of the angle BCA to the sine of the angle DCA.

For, by 22, the co-sine of the angle B is to the sine of the angle
BCA as (the co-sme of the side AC is to the radius ;, that is, by
prop. 22, as) the co-sine of the angle D to the sine of the angle
DCA ; and, by permutation, the co-sine of the angle B is to the
co-sine of the angle D, as the sine of the angle BCA to the sine o£
the angle DCA. Q. E. D.

PROP. XXV. FIG. 17- 18.

The same things remaining, the co-sines of the sides BC, CD,
are proportional to the co-sines of the bases BA, AD.

For, by 21, the co-sine of BC is to the co-sine of BA, as (the co-sine
of AC to the radius; that is, by 21, as) the co-sine of CD is to the
co-sine of AD ; wherefore, by permutation, the co-sines of the sides
BC, CD are proportional to the co-sines of the bases BA, AD. €U
B.D.

PROP. XXVL FIG. 17. 18.

The same construction remaining the sines of the bases BA,
AD are reciprocally proportional to the tangents of the angles
B and D at the base^

SPHERICAL TRIGONOMETRY. 411

For, by 17, the sine of BA is to the radius, as the tangent of AC
to the tangent of the angle B; and by 17, and inversion, the radius
is to the sine of AD, as the tangent of D to the tangent of AC : there-
fore, ex aequo perturbate, the sine of BA is to the sine of AD, as the
tangent of D to the taogent of B.

PROP. XXVn. FIG. 17. 18.

The co-sines of the vertical angles are reciprocally proportional
to the tangents of the sides.

For, by prop. 20, the co-sine of the angle BCA is to the radius as
the tangent of CA is to the tangent of BC; and by the same prop. 20,
and by inversion, the radius is to the co-sine of the angle DCA, as
the tangent of DC to the tangent of CA : therefore, ex aequo pertur-
bate, the co-sine of the angle BCA is to the co-sine of the angle DCA,
as the tangent of DC is to the tangent of BC. Q. E. D.

LEMMA. FIG. 19. 20.

Iir right angled plane triangles, the hypothenuse is to the radi-
us, as the excess of the hypothenuse above either of the sides to
the versed sine of the acute angle adjacent to that side, or as the
sum of the hypothenuse, and either of the sides to the versed sine
of the exterior angle of the triangle.

Let the triangle ABC have a right angle at B ; AC will be to the
radius as the excess of AC above AB, to the versed sine of the angle
A adjacent to AB; or as the sum of AC, AB to the versed sine of the
exterior angle CAK.

With any radius DE, let a circle be described, and from D the
centre, let DF be drawn to the circumference, making the angle EDF
equal to the angle BAC, and from the point F, let FG be drawn per-
pendicular to DE ; let AH, AE be made equal to AC, and DL to DE :
DG therefore is the co-sine of the angle EDF or BAC, and GE its
versed sine: and because of the equiangular triangles ACB, DFG,
AC or AH is to DF or DE, as AB to DG : therefore (19. 5.) AC is to
the radius PE as BH to GE, the versed sine of the angle EDF or
BAC : and since AH is to DE, as AB to DG (12. 5.), AH or AC will
be to the radius DE as KB to LG, the versed sine of the angle LDF
or KAC. a E. D.

PROP. XXVIII. FIG. 21. 22.

Is any spherical triangle, the rectangle contained by the sines
of two sides, is to the square of the radius, as the excess of the
versed sines of the third side or base, and the arch, which is the
excess of the sides, is to the versed sine of the angle opposite to
the base.

419 SPHERICAL TBIGOKOMETaT. ^

Let ABC be a spherical triangle ; the rectangle contained by the
sines of AB, EC vill be to the square of the radius, as the excess of the
Versed sines of the base AC, and of the arch, which is the excess of
Afi, EC, to the versed sine of the angle ABC opposite to the base.

Let D be the centre of the sphere, and let AD, BD, CD be joined,
and let the sines AJE, CF, CG of the arches AB, BC, AC be drawn ;
let the side BC be greater than BA, and let BH be made equal to
BC ; AH will therefore be the excess of the sides BC, BA; let HK be
drawn perpendicular to AD, and since AG is the versed sme of the
base AC, and AK the versed sine of the arch AH, KG is the excess
cfthe versed sines of the base AC, and of the arch AH, which is the
excess of the sides BC, BA : let GL likewise be drawn parallel to KH,
and let it meet FH in L : let CL, DH be joined, and let AD, FH meet
each other in M.

Since therefore in the triangles CDF, HDF, DC, DH are equal, DF
is common, and the angle FDC equal to the angle FDH; because of
the equal arches BC, BH, the base HF will be equal to the base t'C,
and the angle HFD equal to the right angle CFD : the straight line
DF therefore (4. 11.) is at right angles to the plane CFH : wherefore
the plane CFH is at right angles to the plane BDH, which passes
through DF (18. 11.). In like manner, since DG is at right angles
to both GC and GL, DG will be perpendicular to the plane CGL ;
therefore the plane CGL is at right angles to the plane BDH, which
passes through DG : and it was shown, that the plane CFH or CFL
was perpendicular to the same plane BDH ; therefore the common
section of the planes CFL, CGL, viz. the straight line CL is perpen-
dicular to the plane BDA (19. 11.), and therefore CLF is a right an-
gle: in the triangle CFL having the right angle CLF, by the lemma,
CF is to the radius, as LH, the excess, viz. of CF or FH above FL,
is to the versed sine of the angle CFL ; but the angle CFL is the in-
clination of the planes BCD, BAD, since FC, FL are drawn in them
at right angles to the common section BF : the spherical angle ABC
is therefore the same with the angle CFL ^and therefore CF is to the
radius as LH to the versed sine of the spherical angle ABC ; and
since the triangle AED is equiangular (to the triangle MFD, and
therefore) to the triangle MGL, AE will be to the radius of the sphere
AD, as (MG to ML; that is, because of the parallels, as) GK to LH:
the ratio therefore which is compounded of the ratios of AE to the
radius, and of CF to the same radius ; that is, (23. 6.) the ratio of
the rectangle contained by AE, CF to the square of the radius, is the
same with the ratio compounded of the ratio of GK to IM, and the
ratio of LH to the versed sine of the angle ABC ; that is, the same
with the ratio of GK to the versed sine of the angle ABC ; there-
fore, the rectangle contained by AE, CF, the sines of the sides AB,
BC, is to the square of the radius, as GK the excess of the versed
sines AG, AK, of the base AC, and the arch AH, which is the excess
of the sides, to the versed sine of the angle ABC opposite to the
base A-C. Q. E. D.

8PHBRICAL TRIGONOMETRY. 413

PROP. XXIX. FIG. 23.

The rectangle contained by half of the radius, and the excess
of the versed sines of two arches, is equal to the rectangle con-
tained by the sines of half the sum, and half the difference of the
same ^arches.

Lef AB, AC be any two arches, and let AD be made equal to AC
the less ; the arch DB therefore is the sum, and the arch CB the
difference of AC, AB : through E the centre of the circle, let there
be drawn a diameter DEF, and AE joined, and CD likewise perpen-
dicular to it in G ; and let BH be perpendicular to AE, and AH will
be the versed sine of the arch AB, and AG theljversed sine of AC,
and HG the excess of these versed sines : let BD, BC, BF be joined,
and FC also meeting BH in E.

Since therefore BH, CG are parallel, the alternate angles BKC,
KCG will be equal; but KCG is in a semicircle; and therefore a
right angle; therefore BKO is a right angle ; and in the triangles,
DFB, CBK, the angles FDB, BCE, in the same segment are equal,
and FBD, BKC are right angles ; the triangles DFB, CBK are there-
fore equiangular ; wherefore DF is to DB, as BC to CE, ot HG ;
and therefore the rectangle contained by the diameter DF and HG,
is equal to that contained by DB, BC ; wherefore the rectangle con-
tained by a fourth part of the diameter, and HG, is equal to that
contained by the halves of DB, BC : but half the chord DB is the
sine of half the arch DAB, that is, half the sum of the arches AB,
AC; and half the chord of BC is the sine of half the arch BC, which
is the difference of AB, AC. Whence the proposition is manifest.

PROP. XXX. FIG. 19. 24,

The rectangle contained by half of the radius, and the versed
side of any arch, is equal to the square of the sine of half the
same arch.

Let AB be an arch of a circle, C its centre, and AC, CB, BA being
joined ; let AB be bisected in D, and let CD be joined, which will be
perpendicular to BA, and bisect it in E (4. 1.), BE or AE therefwe
is the sine of the arch DB or AD, the half of AB: let BF be perpen-
dicular to AC, and AF will be the versed sine of the arch BA ; but,
because of the similar triangles CAE, BAF, CA is to AE, as AB, that
is, twice AE, to AF; and by halving the antecedents, half of the
radius CA is to AE, the sine of the arch AD, as the same AE, to AF
the versed sine of the arch AB. Wherefore by 16, 6, the proposition
is manifest.

PROP. XXXI. FIG. 26. '

In a spherical triangle, the rectangle contained by the sines of
the two sides, is to the square of the radius, as the rectangle con-
tained by the sine of the arch which is half the sum of the base,

414 srHERICAL TRtGONOMGTRY.

and the excess of the sides, and the sine of .the arch, which is
half the difference of the same, to the square of the sine of half
the angle opposite to the base.

Let ABC be a spherical triangle, of which the two sides are AB,
EC, and base AC, and let the less side BA be produced, so that BD
shall be equal to BC ; AD therefore is the excess of BC, BA ; wid it
is to be shown, that the rectangle contained by the sines of BC, BA
is to the square of the radius, as the rectangle contained by the sine
of half the sum of AC, AD, and the sine of half the difference of the
same AC, AD to the square of the sine of half the angle ABC, oppo-
site to the base AC.

Since by prop. 28, the rectangle contained by the sines of the sides
BC, BA is to the square of the radius, as the excess of the versed
sines of the base AC and AD, to the versed sine of the angle B ; that
is, (1. 6.) as the rectangle contained by half of the radius, and that
excess, to the rectangle contained by half the radius, and the versed
sine of B ; (therefore 29. 30. of this) the rectangle contained by the
sines of the sides BC, BA is to the square of the radius, as the rect-
angle contained by the sine of the arch, which is half the sum of
AC, AD, and the sine of the arch which is half the difference of the
same AC, AD is to the square of the sine of half the angle ABC.

a. E. D.

SPHERICAL TRIGOI704fETRY.

415^

SOLUTION OP THE TWELVE CASES OF OBLIQUE ANGLED SPHERICAL TRI-
ANGLES.

GENERAL PROPOSITION.

In an oblique angled spherical triangle, of the three sides
and three angles, any three being given, the other three may be
founds '

Given.

B, D, and
BC, two an-
gles and a
side oppo-
site one of
them.
Fig. 26,27.

2B, C, and

BC, two an-
gles and
the side
between
them.

BC, CD, &
B.

BC, DB, &
B.

5B, D, and
BC.

6|BC, BD &>
B.

7BC, DC &
B.

Sought.

BD.

CD.

DB.

Co-S, BC : R : : Co-T, B : T, BCA. 19.
Likewise by 24. Co-S, B : S, BCA : : Co-S,
D : S, DCA ; wherefore BCD is the sum or dif-
ference of the angles DCA, BCA according as the
perpendicular CA falls within or without the
triangle BCD ; that is (16. of this,) according as
the angles B, D, are of the same or different af-
fection.

Co-S, BC : R : : Co-T, B : T, BCA. 19. and
also by 24. S, BCA : S, DCA ; : Co-S^ B : Co-
S, D ; and according as the angle BCA is less or
greater than BCD, the perpendicular CA falls
within or without the triangle BCD ; and there-
fore (16. of this,) the angles B, D will be of the
same or different affection.

R : Co-S, B : : T, BC : T, BA. 20. and Co.S,
BC : Co-S, BA : : Co-S, DC : Co-S, DA. 25. and
BD is the sum or difterence of BA, DA.

R : Co-S, B : : T, BC : T, BA. 20. and Co-S,
BA : Co-S, BC : : Co-S, DA : Co-S, DC. 25. and
according as DA, AC are of the same or differ-
ent affection, DC will be less or greater than a
quadrant 14.

R : Co-S< B : : T, BC : T, BA. 20. and
T, D : T, B : : S, BA : S, DA. 26. and BD isthe
sum or difference of BA, DA.

R : Co-S, B : : T, BC : T, BA. 20. and
S, DA : S, BA : : T, B : T, D ; and according as
BD is greater or less than BA, the angles B, D
are of the same or different affection. 16.

Co-S, BC : R : : Co-T, B : T, BCA. 19. and
T, DC : T, BC : : Co-S, BCA : Co-S, DCA. 27.
the sum or difference of the angles BCA, DCA is
equal to the sngie BCD.

416

SPHERICAL TftlGpNOMETRY.

B,CtAd
BC.

10 g

, D and
BC.

Given;

BC, CD
andB.

BC, BA,
AC.

Pig. 25.

Ay Sf C.

Fig. 7.

Sought

J^.

DC.

Co-S, BC : R : ; CoT, B : T, BCA. 19. also by
27. C<>S|CDA : Co-S, BCA: : T, BC: T.DC.27.
if DCA and B be of the same affection; that is
(13.) if AD and CA be similar, DC will be leas
than a qfiadranL 14. uid if AD, CA be not of the
same anection, DC is greater than a quadrant 14.

S, CD : S, B : : S, BC : S, D.

S, D : Si BC : : S, B : S, DC.

S, AB X S, BC : R^ : : S, AC + AD X S

AC— AD : Sq. ABC. See Pig. 25. AD being

The sides.

the difference of the side^ BC,<BA.

See Pig. 7.

In the trianffle DEP, DE, EP, PD are respec-
tively the supplements of the measures of the siven
angles B» A, C, in the triangle BAC; the siaes of
the triangle D£F are therefore given, and by the
preceding case the angles D> E, P may be found,
and the sides BC, BA, AC are the supplements of
the measures of ^ese angles.

The 8d, 5th, 7th, 9th, 10th cases, which afe commonly called
ambiguouSf admit of two solutions, either of which will answer, the
conditions required ; for,- in these cases, the measure of the angle
or side sought, may be either greater or less than a quadrant, and
the two solutions will be supplements to each other. (Cor. to def.
4. 6. PL Tr.)

If from any of the angles of an oblique ang]^ spherical triangle,
a perpendicular arch be drawn upon the opposite side, most of the
cases of oblique angled triangles may be reserved by means of Na-
pier's rules. ' y

FINIS.

Fh/,1

IJr/, .^.

»/.

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